# 2.E: Computing Derivatives (Exercises)

## 2.1: Elementary Derivative Rules

1. Let f and g be differentiable functions for which the following information is known: f (2) = 5, g(2) = −3, f 0 (2) = −1/2, g 0 (2) = 2.

(a) Let h be the new function defined by the rule h(x) = 3 f (x) − 4g(x). Determine h(2) and h 0 (2).

(b) Find an equation for the tangent line to y = h(x) at the point (2, h(2)).

(c) Let p be the function defined by the rule p(x) = −2 f (x)+ 1 2 g(x). Is p increasing, decreasing, or neither at a = 2? Why?

(d) Estimate the value of p(2.03) by using the local linearization of p at the point (2, p(2)).

2. Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.1. Use the graphs to answer the following questions.

Figure 2.1: The graphs of p (in blue) and q (in green).

(a) At what values of x is p not differentiable? At what values of x is q not differentiable? Why?

(b) Let r(x) = p(x) + 2q(x). At what values of x is r not differentiable? Why?

(c) Determine r 0 (−2) and r 0 (0).

(d) Find an equation for the tangent line to y = r(x) at the point (2,r(2)). 95

3. Consider the functions r(t) = t t and s(t) = arccos(t), for which you are given the facts that r 0 (t) = t t (ln(t) + 1) and s 0 (t) = − 1 √ 1−t 2 . Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain 0 < t < 1.

(a) Let w(t) = 3t t − 2 arccos(t). Determine w 0 (t).

(b) Find an equation for the tangent line to y = w(t) at the point ( 1 2 , w( 1 2 )).

(c) Let v(t) = t t + arccos(t). Is v increasing or decreasing at the instant t = 1 2 ? Why?

4. Let f (x) = a x . The goal of this problem is to explore how the value of a affects the derivative of f (x), without assuming we know the rule for d dx [a x ] that we have stated and used in earlier work in this section.

(a) Use the limit definition of the derivative to show that f 0 (x) = lim h→0 a x · a h − a x h .

(b) Explain why it is also true that f 0 (x) = a x · lim h→0 a h − 1 h .

(c) Use computing technology and small values of h to estimate the value of L = lim h→0 a h − 1 h when a = 2. Do likewise when a = 3.

(d) Note that it would be ideal if the value of the limit L was 1, for then f would be a particularly special function: its derivative would be simply a x , which would mean that its derivative is itself. By experimenting with different values of a between 2 and 3, try to find a value for a for which L = lim h→0 a h − 1 h = 1.

(e) Compute ln(2) and ln(3). What does your work in (b) and (c) suggest is true about d dx [2 x ] and d dx [3 x ]? (f) How do your investigations in (d) lead to a particularly important fact about the function f (x) = e x ?

## 2.2: The Sine and Cosine Function

1. Suppose that V(t) = 24 · 1.07t + 6 sin(t) represents the value of a person’s investment portfolio in thousands of dollars in year t, where t = 0 corresponds to January 1, 2010.

(a) At what instantaneous rate is the portfolio’s value changing on January 1, 2012? Include units on your answer.

(b) Determine the value of V 00(2). What are the units on this quantity and what does it tell you about how the portfolio’s value is changing?

(c) On the interval 0 ≤ t ≤ 20, graph the function V(t) = 24 · 1.07t + 6 sin(t) and describe its behavior in the context of the problem. Then, compare the graphs of the functions A(t) = 24 · 1.07t and V(t) = 24 · 1.07t + 6 sin(t), as well as the graphs of their derivatives A 0 (t) and V 0 (t). What is the impact of the term 6 sin(t) on the behavior of the function V(t)? 2. Let f (x) = 3 cos(x) − 2 sin(x) + 6.

(a) Determine the exact slope of the tangent line to y = f (x) at the point where a = π 4 .

(b) Determine the tangent line approximation to y = f (x) at the point where a = π.

(c) At the point where a = π 2 , is f increasing, decreasing, or neither?

(d) At the point where a = 3π 2 , does the tangent line to y = f (x) lie above the curve, below the curve, or neither? How can you answer this question without even graphing the function or the tangent line? 101

3. In this exercise, we explore how the limit definition of the derivative more formally shows that d dx [sin(x)] = cos(x). Letting f (x) = sin(x), note that the limit definition of the derivative tells us that f 0 (x) = lim h→0 sin(x + h) − sin(x) h .

(a) Recall the trigonometric identity for the sine of a sum of angles α and β: sin(α + β) = sin(α) cos(β) + cos(α)sin(β). Use this identity and some algebra to show that f 0 (x) = lim h→0 sin(x)(cos(h) − 1) + cos(x)sin(h) h .

(b) Next, note that as h changes, x remains constant. Explain why it therefore makes sense to say that f 0 (x) = sin(x) · lim h→0 cos(h) − 1 h + cos(x) · lim h→0 sin(h) h .

(c) Finally, use small values of h to estimate the values of the two limits in (c): lim h→0 cos(h) − 1 h and lim h→0 sin(h) h .

(d) What do your results in (c) thus tell you about f 0 (x)?

(e) By emulating the steps taken above, use the limit definition of the derivative to argue convincingly that d dx [cos(x)] = − sin(x).

## 2.3: The Product and Quotient Rules

1. Let f and g be differentiable functions for which the following information is known: f (2) = 5, g(2) = −3, f 0 (2) = −1/2, g 0 (2) = 2.

(a) Let h be the new function defined by the rule h(x) = g(x) · f (x). Determine h(2) and h 0 (2).

(b) Find an equation for the tangent line to y = h(x) at the point (2, h(2)) (where h is the function defined in (a)).

(c) Let r be the function defined by the rule r(x) = g(x) f (x) . Is r increasing, decreasing, or neither at a = 2? Why? 111

(d) Estimate the value of r(2.06) (where r is the function defined in (c)) by using the local linearization of r at the point (2,r(2)).

2. We restrict our interest in both functions to the domain 0 < t < 1.

(a) Let w(t) = t t arccos(t). Determine w 0 (t).

(b) Find an equation for the tangent line to y = w(t) at the point ( 1 2 , w( 1 2 )).

(c) Let v(t) = t t arccos(t) . Is v increasing or decreasing at the instant t = 1 2 ? Why?

3. Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.5. Use the graphs to answer the following questions.

Figure 2.5: The graphs of p (in blue) and q (in green).

(a) Let r(x) = p(x) · q(x). Determine r 0 (−2) and r 0 (0).

(b) Are there values of x for which r 0 (x) does not exist? If so, which values, and why?

(c) Find an equation for the tangent line to y = r(x) at the point (2,r(2)).

(d) Let z(x) = q(x) p(x) . Determine z 0 (0) and z 0 (2).

(e) Are there values of x for which z 0 (x) does not exist? If so, which values, and why?

4. A farmer with large land holdings has historically grown a wide variety of crops. With the price of ethanol fuel rising, he decides that it would be prudent to devote more and more of his acreage to producing corn. As he grows more and more corn, he learns efficiencies that increase his yield per acre. In the present year, he used 7000 acres of 112 his land to grow corn, and that land had an average yield of 170 bushels per acre. At the current time, he plans to increase his number of acres devoted to growing corn at a rate of 600 acres/year, and he expects that right now his average yield is increasing at a rate of 8 bushels per acre per year. Use this information to answer the following questions.

(a) Say that the present year is t = 0, that A(t) denotes the number of acres the farmer devotes to growing corn in year t, Y(t) represents the average yield in year t (measured in bushels per acre), and C(t) is the total number of bushels of corn the farmer produces. What is the formula for C(t) in terms of A(t) and Y(t)? Why?

(b) What is the value of C(0)? What does it measure?

(c) Write an expression for C 0 (t) in terms of A(t), A 0 (t), Y(t), and Y 0 (t). Explain your thinking.

(d) What is the value of C 0 (0)? What does it measure?

(e) Based on the given information and your work above, estimate the value of C(1).

5. Let f (v) be the gas consumption (in liters/km) of a car going at velocity v (in km/hour). In other words, f (v) tells you how many liters of gas the car uses to go one kilometer if it is traveling at v kilometers per hour. In addition, suppose that f (80) = 0.05 and f 0 (80) = 0.0004.

(a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. What is the relationship between f (v) and g(v)? Hence find g(80) and g 0 (80).

(b) Let h(v) be the gas consumption in liters per hour of a car going at velocity v. In other words, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the algebraic relationship between h(v) and f (v)? Hence find h(80) and h 0 (80).

(c) How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.

## 2.4: Derivatives of Other Trigonometric Functions

1. An object moving vertically has its height at time t (measured in feet, with time in seconds) given by the function h(t) = 3 + 2 cos(t) 1.2 t .

(a) What is the object’s instantaneous velocity when t = 2?

(b) What is the object’s acceleration at the instant t = 2?

(c) Describe in everyday language the behavior of the object at the instant t = 2.

2. Let f (x) = sin(x) cot(x).

(a) Use the product rule to find f 0 (x). 118

(b) True or false: for all real numbers x, f (x) = cos(x).

(c) Explain why the function that you found in (a) is almost the opposite of the sine function, but not quite. (Hint: convert all of the trigonometric functions in (a) to sines and cosines, and work to simplify. Think carefully about the domain of f and the domain of f 0 .)

3. Let p(z) be given by the rule p(z) = z tan(z) z 2 sec(z) + 1 + 3e z + 1.

(a) Determine p 0 (z).

(b) Find an equation for the tangent line to p at the point where z = 0.

(c) At z = 0, is p increasing, decreasing, or neither? Why?

## 2.5: The Chain Rule

1. Consider the basic functions f (x) = x 3 and g(x) = sin(x).

(a) Let h(x) = f (g(x)). Find the exact instantaneous rate of change of h at the point where x = π 4 .

(b) Which function is changing most rapidly at x = 0.25: h(x) = f (g(x)) or r(x) = g( f (x))? Why?

(c) Let h(x) = f (g(x)) and r(x) = g( f (x)). Which of these functions has a derivative that is periodic? Why?

2. Let u(x) be a differentiable function. For each of the following functions, determine the derivative. Each response will involve u and/or u 0 .

(a) p(x) = e u(x)

(b) q(x) = u(e x )

(c) r(x) = cot(u(x))

(d) s(x) = u(cot(x))

(e) a(x) = u(x 4 )

(f) b(x) = u 4 (x)

3. Let functions p and q be the piecewise linear functions given by their respective graphs in Figure 2.7. Use the graphs to answer the following questions.

(a) Let C(x) = p(q(x)). Determine C 0 (0) and C 0 (3).

(b) Find a value of x for which C 0 (x) does not exist. Explain your thinking.

(c) Let Y(x) = q(q(x)) and Z(x) = q(p(x)). Determine Y 0 (−2) and Z 0 (0). 128

Figure 2.7: The graphs of p (in blue) and q (in green).

4. If a spherical tank of radius 4 feet has h feet of water present in the tank, then the volume of water in the tank is given by the formula V = π 3 h 2 (12 − h).

(a) At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant h = 1? What are the units on this quantity?

(b) Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time t is given by the rule h(t) = sin(πt) + 1, where t is measured in hours (and h is still measured in feet). At what rate is the height of the water changing with respect to time at the instant t = 2?

(c) Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant t = 2?

(d) What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.

## 2.6: Derivatives of Inverse Functions

1. Determine the derivative of each of the following functions. Use proper notation and clearly identify the derivative rules you use.

(a) f (x) = ln(2 arctan(x) + 3 arcsin(x) + 5)

(b) r(z) = arctan(ln(arcsin(z)))

(c) q(t) = arctan2 (3t) arcsin4 (7t)

(d) g(v) = ln arctan(v) arcsin(v) + v 2 !

2. Consider the graph of y = f (x) provided in Figure 2.13 and use it to answer the following questions.

(a) Use the provided graph to estimate the value of f 0 (1).

(b) Sketch an approximate graph of y = f −1 (x). Label at least three distinct points on the graph that correspond to three points on the graph of f .

(c) Based on your work in (a), what is the value of ( f −1 ) 0 (−1)? Why?

3. Let f (x) = 1 4 x 3 + 4.

(a) Sketch a graph of y = f (x) and explain why f is an invertible function.

Figure 2.13: A function y = f (x) for use in Exercise 2.

(b) Let g be the inverse of f and determine a formula for g.

(c) Compute f 0 (x), g 0 (x), f 0 (2), and g 0 (6). What is the special relationship between f 0 (2) and g 0 (6)? Why?

4. Let h(x) = x + sin(x).

(a) Sketch a graph of y = h(x) and explain why h must be invertible.

(b) Explain why it does not appear to be algebraically possible to determine a formula for h −1 .

(c) Observe that the point ( π 2 , π 2 + 1) lies on the graph of y = h(x). Determine the value of (h −1 ) 0 ( π 2 + 1).

## 2.7: Derivatives of Functions Given Implicitely

1. Consider the curve given by the equation 2y 3 + y 2 − y 5 = x 4 − 2x 3 + x 2 . Find all points at which the tangent line to the curve is horizontal or vertical. Be sure to use a graphing utility to plot this implicit curve and to visually check the results of algebraic reasoning that you use to determine where the tangent lines are horizontal and vertical.

2. For the curve given by the equation sin(x + y) + cos(x − y) = 1, find the equation of the tangent line to the curve at the point ( π 2 , π 2 ).

3. Implicit differentiation enables us a different perspective from which to see why the rule d dx [a x ] = a x ln(a) holds, if we assume that d dx [ln(x)] = 1 x . This exercise leads you through the key steps to do so.

(a) Let y = a x . Rewrite this equation using the natural logarithm function to write x in terms of y (and the constant a).

(b) Differentiate both sides of the equation you found in (a) with respect to x, keeping in mind that y is implicitly a function of x.

(c) Solve the equation you found in (b) for dy dx , and then use the definition of y to write dy dx solely in terms of x. What have you found?

## 2.8: Using Derivatives to Evaluate Limits

1. Let f and g be differentiable functions about which the following information is known: f (3) = g(3) = 0, f 0 (3) = g 0 (3) = 0, f 00(3) = −2, and g 00(3) = 1. Let a new function h be given by the rule h(x) = f (x) g(x) . On the same set of axes, sketch possible graphs of f and g near x = 3, and use the provided information to determine the value of lim x→3 h(x). Provide explanation to support your conclusion.

2. Find all vertical and horizontal asymptotes of the function R(x) = 3(x − a)(x − b) 5(x − a)(x − c) , where a, b, and c are distinct, arbitrary constants. In addition, state all values of x for which R is not continuous. Sketch a possible graph of R, clearly labeling the values of a, b, and c.

3. Consider the function g(x) = x 2x , which is defined for all x > 0. Observe that limx→0 + g(x) is indeterminate due to its form of 0 0 . (Think about how we know that 0 k = 0 for all k > 0, while b 0 = 1 for all b , 0, but that neither rule can apply to 0 0 .)

(a) Let h(x) = ln(g(x)). Explain why h(x) = 2x ln(x).

(b) Next, explain why it is equivalent to write h(x) = 2 ln(x) 1 x .

(c) Use L’Hopital’s Rule and your work in (b) to compute limx→0 + h(x).

(d) Based on the value of limx→0 + h(x), determine limx→0 + g(x).

4. Recall we say that function g dominates function f provided that limx→∞ f (x) = ∞, limx→∞ g(x) = ∞, and limx→∞ f (x) g(x) = 0.

(a) Which function dominates the other: ln(x) or √ x?

(b) Which function dominates the other: ln(x) or n √ x? (n can be any positive integer)

(c) Explain why e x will dominate any polynomial function.

(d) Explain why x n will dominate ln(x) for any positive integer n. 160

(e) Give any example of two nonlinear functions such that neither dominates the other

## CLP-1 Differential Calculus

At this point we could try to start working out how derivatives interact with arithmetic and make an “Arithmetic of derivatives” theorem just like the one we saw for limits (Theorem 1.4.3). We will get there shortly, but before that it is important that we become more comfortable with computing derivatives using limits and then understanding what the derivative actually means. So — more examples.

###### Example 2.2.9 (diff<>sqrt)

Compute the derivative, (f'(a) ext<,>) of the function (f(x)=sqrt) at the point (x=a) for any (a gt 0 ext<.>)

So again we start with the definition of derivative and go from there:

We should think about the domain of (f') here — that is, for which values of (a) is (f'(a)) defined? The original function (f(x)) was defined for all (x geq 0 ext<,>) however the derivative (f'(a)=frac<1><2sqrt>) is undefined at (a = 0 ext<.>)

If we draw a careful picture of (sqrt) around (x=0) we can see why this has to be the case. The figure below shows three different tangent lines to the graph of (y=f(x)=sqrt ext<.>) As the point of tangency moves closer and closer to the origin, the tangent line gets steeper and steeper. The slope of the tangent line at (ig(a,sqrtig)) blows up as (a o 0 ext<.>)

###### Example 2.2.10 (diff<> left\)

Compute the derivative, (f'(a) ext<,>) of the function (f(x)=|x|) at the point (x=a ext<.>)

We should start this example by recalling the definition of (|x|) (we saw this back in Example 1.5.6):

It is definitely not just “chop off the minus sign”.

Since (x gt 0) and we are interested in the behaviour of this function as (h o 0) we can assume (h) is much smaller than (x ext<.>) This means (x+h gt 0) and so (|x+h|=x+h ext<.>)

Since (x lt 0) and we are interested in the behaviour of this function as (h o 0) we can assume (h) is much smaller than (x ext<.>) This means (x+h lt 0) and so (|x+h|=-(x+h) ext<.>)

Are you working to calculate derivatives in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

### Trigonometric Derivatives

Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

### Constant Factor Rule

Constants come out in front of the derivative, unaffected:
$dfracleft[c f(x) ight] = c dfracf(x)$

For example, $dfracleft(4x^3 ight) = 4 dfracleft(x^3 ight) =, …$

### Sum of Functions Rule

The derivative of a sum is the sum of the derivatives:
$dfrac left[f(x) + g(x) ight] = dfracf(x) + dfracg(x)$

For example, $dfracleft(x^2 + cos x ight) = dfracleft( x^2 ight) + dfrac(cos x) = , …$

### IV. Quotient Rule for Derivatives

Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing

“lo d-hi minus hi d-lo over lo-lo”

Two specific cases you’ll quickly remember:
$dfrac ext <(constant)>= 0$
$dfrac(x) = 1$

Differentiate $f(x) = sqrt$.
Recall $sqrt[n] = x^<1/n>.$
Differentiate $f(x) = dfrac<5>$.
Recall $dfrac<1> = x^<-n>.$

“lo d-hi minus hi d-lo over lo-lo”

Have a question, suggestion, or item you’d like us to include? Please let us know in the Comments section below!

Want access to all of our Calculus problems and solutions? Buy full access now — it’s quick and easy!

## Math Insight

Let $f(x,y)=x^2+y^2$. Find $Df(1,2)$ and the equation for the tangent plane at $(x,y)=(1,2)$. Find the linear approximation to $f(x,y)$ at $(x,y)=(1,2)$.

Solution: egin pdiff(x,y) &= 2x pdiff(1,2) &= 2 pdiff(x,y) &= 2y pdiff(1,2) &= 4 end So $Df(1,2)=left[ 2 4 ight]$.

Since both partial derivatives $pdiff(x,y)$ and $pdiff(x,y)$ are continuous functions, we know that $f(x,y)$ is differentiable. Therefore, $Df(1,2)$ is the derivative of $f$, and the function has a tangent plane there.

To calculate the equation of the tangent plane, the only additional calculation is the value of $f$ at $(x,y)=(1,2)$, which is $f(1,2) = 1^2+2^2=5$. The equation for the tangent plane is egin z &= f(1,2)+pdiff(1,2)(x-1) + pdiff(1,2)(y-2) &= 5 + 2(x-1) + 4(y-2) end

For a scalar-valued function of two variables such as $f(x,y)$, the tangent plane is the linear approximation. We can write the linear approximation as egin L(x,y) = 5 + 2(x-1) + 4(y-2). end

#### Example 1'

If looked at the point $(2,3)$, what changes?

Solution: The partial derivatives change, so the derivative becomes egin pdiff(2,3) &= 4 pdiff(2,3) &= 6 Df(2,3) &= left[ 4 6 ight]. end The equation for the tangent plane, i.e., the linear approximation, becomes egin z &= L(x,y) = f(2,3)+pdiff(2,3)(x-2) + pdiff(2,3)(y-3) &= 13 + 4(x-2) + 6(y-3) end

#### Example 2

Find the derivative of egin vc(x,y,z)=(x^2y^2z,y+sin z) end at the point $(1,2,0)$.

Solution: $vc: R^3 ightarrow R^2$ (confused?), so the derivative (assuming the function is differentiable) is the $2 imes 3$ matrix of partial derivatives. The partial derivatives of the matrix are egin pdiff &= 2xy^2z pdiff &= 2x^2yz pdiff &= x^2y^2 pdiff &= 0 pdiff &= 1 pdiff &= cos z. end Since all these functions are continuous, $vc$ is differentiable.

For any point $(x,y,z)=(a,b,c)$, the derivative is egin Dvc(a,b,c) = left[ egin 2ab^2c & 2a^2bc & a^2b^2 0 & 1 & cos c end ight]. end At $(1,2,0)$, the derivative is egin Dvc(1,2,0) = left[ egin 0 & 0 & 4 0 & 1 & 1 end ight]. end

#### Example 3

For the function of example 2, calculate the linear approximation to $vc$ at the point $(1,2,0)$.

Solution: We've already calculated almost everything we need. We also need the value of the function at (1,2,0): egin vc(1,2,0) = &=((1^2) (2^2) 0, 2+sin 0) &= (0,2) end Then, the linear approximation to $vc$ at (1,2,0) is A linear approximation to $vc$ is egin L(x,y,z) & = vc(1,2,0) + Dvc(1,2,0) (x-1, y-2, z) &= left[ egin 0 2 end ight] + left[ egin 0 & 0 & 4 0 & 1 & 1 end ight] left[ egin x-1 y-2 z end ight] &= left[ egin 0 2 end ight] + left[ egin 4z y-2+z end ight] &=(4z, y+z) end

#### Example 4

Use the linear approximation of $vc(x,y,z)$ from Example 3 to approximate the value of $vc$ at the point $(1.1,1.9,0.1)$.

Solution: The above linear approximation at $(x,y,z) = (1.1,1.9,0.1)$ is egin L(1.1,1.9,0.1) &= (4(0.1), 1.9+0.1) & = (0.4, 2.0) end

Note that $(1.1,1.9,0.1)$ is very close to $(1,2,0)$, which is the point around which we computed the linear approximation. So, we expect this linear approximation to be close to the true value of $vc$ at $(1.1,1.9,0.1)$. Let's compare the above answer to the actual value of $vc$ at $(1.1,1.9,0.1)$: egin vc(1.1,1.9,0.1) &= ((1.1)^2(1.9)^2(0.1), 1.9+sin (0.1)) &approx (0.4368,1.9998). end In this case, the approximation is quite close.

## Proof

If and , then . Differentiating both sides of this equation results in the equation

.

Solving for yields

.

Finally, we substitute to obtain

.

We may also derive this result by applying the inverse function theorem, as follows. Since is the inverse of , by applying the inverse function theorem we have

.

Using this result and applying the chain rule to yields

The graph of and its derivative are shown in (Figure).

Figure 3. The function is increasing on . Its derivative is greater than zero on .

### Taking a Derivative of a Natural Logarithm

Find the derivative of .

### Using Properties of Logarithms in a Derivative

Find the derivative of .

#### Solution

At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.

Differentiate: .

#### Solution

Use a property of logarithms to simplify before taking the derivative.

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of and for .

### Derivatives of General Exponential and Logarithmic Functions

Let , and let be a differentiable function.

If , then

.

More generally, if , then for all values of for which ,

. .

More generally, if , then

.

## Linear Algebra for Team-Based Inquiry Learning

We've seen that row reducing all the way into RREF gives us a method of computing determinants.

However, we learned in module E that this can be tedious for large matrices. Thus, we will try to figure out how to turn the determinant of a larger matrix into the determinant of a smaller matrix.

###### Activity 5.2.2 .

The following image illustrates the transformation of the unit cube by the matrix (left[egin 1 & 1 & 0 1 & 3 & 1 0 & 0 & 1end ight] ext<.>)

Recall that for this solid (V=Bh ext<,>) where (h) is the height of the solid and (B) is the area of its parallelogram base. So what must its volume be?

(displaystyle det left[egin 1 & 1 1 & 3 end ight])

(displaystyle det left[egin 1 & 0 3 & 1 end ight])

(displaystyle det left[egin 1 & 1 0 & 1 end ight])

(displaystyle det left[egin 1 & 3 0 & 0 end ight])

###### Fact 5.2.4 .

If row (i) contains all zeros except for a (1) on the main (upper-left to lower-right) diagonal, then both column and row (i) may be removed without changing the value of the determinant.

Since row and column operations affect the determinants in the same way, the same technique works for a column of all zeros except for a (1) on the main diagonal.

###### Activity 5.2.5 .

Remove an appropriate row and column of (det left[egin 1 & 0 & 0 1 & 5 & 12 3 & 2 & -1 end ight]) to simplify the determinant to a (2 imes 2) determinant.

###### Activity 5.2.6 .

Simplify (det left[egin 0 & 3 & -2 2 & 5 & 12 0 & 2 & -1 end ight]) to a multiple of a (2 imes 2) determinant by first doing the following:

Factor out a (2) from a column.

Swap rows or columns to put a (1) on the main diagonal.

###### Activity 5.2.7 .

Simplify (det left[egin 4 & -2 & 2 3 & 1 & 4 1 & -1 & 3end ight]) to a multiple of a (2 imes 2) determinant by first doing the following:

Use row/column operations to create two zeroes in the same row or column.

Factor/swap as needed to get a row/column of all zeroes except a (1) on the main diagonal.

###### Observation 5.2.8 .

Using row/column operations, you can introduce zeros and reduce dimension to whittle down the determinant of a large matrix to a determinant of a smaller matrix.

###### Activity 5.2.9 .

as a multiple of a determinant of a (3 imes3) matrix.

###### Activity 5.2.10 .

Compute (detleft[egin 2 & 3 & 5 & 0 0 & 3 & 2 & 0 1 & 2 & 0 & 3 -1 & -1 & 2 & 2 end ight]) by using any combination of row/column operations.

###### Observation 5.2.11 .

Another option is to take advantage of the fact that the determinant is linear in each row or column. This approach is called or .

###### Observation 5.2.12 .

Applying Laplace expansion to a (2 imes 2) matrix yields a short formula you may have seen:

There are formulas for the determinants of larger matrices, but they can be pretty tedious to use. For example, writing out a formula for a (4 imes 4) determinant would require 24 different terms!

So this is why we either use Laplace expansion or row/column operations directly.

###### Activity 5.2.13 .

Based on the previous activities, which technique is easier for computing determinants?

Using row/column operations.

Some other technique (be prepared to describe it).

###### Activity 5.2.14 .

Use your preferred technique to compute (detleft[egin 4 & -3 & 0 & 0 1 & -3 & 2 & -1 3 & 2 & 0 & 3 0 & -3 & 2 & -2 end ight] ext<.>)

## 2.E: Computing Derivatives (Exercises)

We have seen in the previous page how the derivative is defined: For a function f ( x ), its derivative at x = a is defined by

Let us give some examples.

Example 1. Let us start with the function f ( x ) = x 2 . We have

What about the derivative of f ( x ) = x n . Similar calculations, using the binomial expansion for ( x + y ) n (Pascal's Triangle), yield

Example 2. Consider the function f ( x )=1/ x for . We have

Have you noticed? The algebraic trick in both of the examples above has been to factor out " h " in the numerator, so that we can cancel it with the " h " in the denominator! This is what you try to do whenever you are asked to compute a derivative using the limit definition.

You may believe that every function has a derivative. Unfortunately that is not the case.

Example 3. Let us discuss the derivative of f ( x ) = | x | at 0. We have

which implies that f '(0) does not exist.

Remark. This example is interesting. Even though the derivative at the point does not exist, the right and the left limit of the ratio do exist. In fact, if we use the slope-interpretation of the derivative we see that this means that the graph has two lines close to it at the point under consideration. They could be seen as "half-tangents". See Picture.

So let's push it a little bit more and ask whether a function always has a tangent or half-tangents at any point. That is not the case either.

Example 4. Let us consider the function for , with f (0) = 0. We have

Recall that the function has no limit when x goes to 0. So the function has no derivative and no half-derivatives as well at x =0.

Example 5. Consider the function . Then we have

then f '(0) does not exist. But observe that the graph as a geometric figure has a tangent -- albeit vertical:

In fact, the way the concept of the tangent line was introduced is based on the notion of slope. You already know that vertical lines do not have slopes. So we say that the derivative does not exist whenever the tangent line is vertical. Nevertheless keep in mind that when the limit giving the derivative is then the function has a vertical tangent line at the point.

It can be quite laborious (or impossible) to compute the derivative by hand as we have done so far. In the next pages we will show how techniques of differentiation help bypass the limit calculations and make our life much easier.

Exercise 1. Find the derivative of

Exercise 2. Discuss the differentiability of

Exercise 3. We say that the graph of f ( x ) has a cusp at ( a , f ( a )), if f ( x ) is continuous at a and if the following two conditions hold: 1. as from one side (left or right) 2. as from the other side.

Determine whether f ( x ) = x 4/3 and g ( x ) = x 3/5 have a cusp at (0,0).

Exercise 4. Show that if f '( a ) exists, then we have

Exercise 5. A spherical balloon is being inflated. Find the rate at which its volume V is changing with respect to the radius.

With functions of one variable we were interested in places where the derivative is zero, since they made candidate points for the maximum or minimum of a function. If the derivative is not zero, we have a direction that is downhill and moving a little in that direction gives a lower value of the function. Similarly, with functions of two variables we can only find a minimum or maximum for a function if both partial derivatives are 0 at the same time. Such points are called critical points.

The point ((a,b)) is a for the multivariable function (f(x,y) ext<,>) if both partial derivatives are 0 at the same time.

###### Example 6.3.1 . Finding a Local Minimum of a Function.

Use the partial derivatives of (f(x,y)=x^2+ 2xy+3y^2-4x-3y) to find the minimum of the graph.

: In the previous section, we already computed

We need to find the places where both partial derivatives are 0. With this simple system, I can solve this system algebraically and find the only critical point is ((9/4, -1/4) ext<.>)

Subtract the equations to eliminate (x ext<.>)

Substitute back and solve for (x ext<.>)

: However, if the partials are more complicated, I will want to find the critical points another way. I can find the point with Solver.

To get solver to set both partials to 0 at the same time, I ask it to solve for (f_y=0 ext<,>) while setting (f_x=0) as a constraint. Make sure to uncheck the box that makes unconstrained variables non-negative.

This finds our critical point within our error tolerance.

: We can also use Wolfram|Alpha to find the solution to our system of equations.

We thus get a critical point at (9/4,-1/4) with any of the three methods of solving for both partial derivatives being zero at the same time. Once we have a critical point we want to determine if it is a maximum, minimum, or something else. The easiest way is to look at the graph near the critical point.

It is clear from the graph that this critical point is a local minimum.

It is easy to see that (f(x,y)=x^2+y^2) has a critical point at ((0,0)) and that that point is a minimum for the function. Similarly, (f(x,y)=-x^2-y^2) has a critical point at ((0,0)) and that that point is a maximum for the function. For some functions, like (f(x,y)=x^2-y^2 ext<,>) which has a critical point at ((0,0) ext<,>) we can have a maximum in one direction and a minimum in another direction. Such a point is called a . We note that we can have a saddle point even if the (x) and (y) slice curves both indicate a minimum.

###### Example 6.3.2 . A Saddle Point at a Minimum on Both Axes.

Show that (f(x,y)=x^2-3xy+y^2) has a critical point at ((0,0) ext<,>) which is a minimum of both slice curves, but is not a local minimum.

Solution: We look at the two partial derivatives, and notice they are both zero at the origin.

We then see that both slice curves are parabolas that bend up, with a minimum at 0.

However if we take the slice with (x=y ext<,>) we get a parabola bending down, so we don’t have a minimum.

Looking at the graph, we see that this graph does not have a minimum.

With only first derivatives, we can just find the critical points. To check if a critical point is maximum, a minimum, or a saddle point, using only the first derivative, the best method is to look at a graph to determine the kind of critical point. For some applications we want to categorize the critical points symbolically.

With functions of one variable we used the second derivative to test if a critical point was a maximum or minimum. In the two variable case we need to define the second derivatives and use them to define the discriminant of a function to test if a critical point is a minimum, maximum, or saddle point. We first need to define second partial derivatives.

Note that (f_) is simply the old second derivative of the curve (f(x,y_0)) and (f_) is simply the old second derivative of the curve (f(x_0,y) ext<.>) For curves with continuous second partial derivatives, the mixed partials, (f_) and (f_) are the same.

###### Example 6.3.3 . Finding Second Partial Derivatives.

Find the second partial derivatives of

Solution: We start by computing the first partial derivatives.

Then we compute the second partial derivatives.

As expected, the mixed partials are the same.

To test if a critical point is a maximum, minimum, or saddle point we compute the discriminant of the function.

###### Example 6.3.4 . Finding the Discriminant of a Function.

Solution: We have already computed the second partial derivatives.

Substituting into the formula

: Let ((a,b)) be a critical point of (f(x,y) ext<.>)

If (D(a,b)>0) and (f_ (a,b)>0) then ((a,b)) is a local minimum of (f(x,y) ext<.>)

If (D(a,b)>0 ) and (f_ (a,b)lt 0) then ((a,b)) is a local maximum of (f(x,y) ext<.>)

If (D(a,b)lt 0) then ((a,b)) is a saddle point of (f(x,y) ext<.>)

If (D(a,b)=0) we do not have enough information to classify the point.

###### Example 6.3.5 . Using the Discriminant to Classify Critical Points.

Based on the information given, classify each of the following points as a local maximum, local minimum, saddle point, not a critical point, or not enough information to classify.

 p () () (f_) (f_) (f_) A 0 0 0 0 1 B 0 1 3 2 4 C 1 0 0 2 3 D 0 0 1 2 0 E 0 0 -1 2 3 F 0 0 -3 1 -2 G 0 0 3 3 3

Solution: We need to compute the discriminant and apply the test.

 p (f_x) (f_y) (f_) (f_) (f_) Discriminant Classification A 0 0 0 0 1 0 Not enough information B 0 1 3 2 4 8 Not a critical point C 1 0 0 2 3 -4 Not a critical point D 0 0 1 2 0 -4 Saddle point E 0 0 -1 2 3 -7 Saddle point F 0 0 -3 1 -2 5 Maximum G 0 0 3 3 3 0 Not enough information
###### Example 6.3.6 . Finding and Classifying Critical Points.

Let (f(x,y)=x^3-3x+y^3-3y^2 ext<.>) Find the critical points and classify them using the discriminant.

Solution: We start by computing the first partial derivatives.

Then we compute the second partial derivatives and the discriminant.

We have critical points when both first partials are 0, so at ((1,2) ext<,>) ((-1,2) ext<,>) ((1,0) ext<,>) and ((-1,0) ext<.>)

At ((1,2) ext<,>) both (D) and (f_) are positive, so we have a local minimum.

At ((-1,2)) and ((1,0) ext<,>) (D) is negative, so we have a saddle point.

At (((-1,0) ext<,>) (D) is positive and (f_) is negative, so we have a local maximum.

### Exercises Exercises: Critical Points and Extrema Problems

For exercises 1-6, for the given functions and region:

Find the partial derivatives of the original function.

Find any critical points in the region.

Produce a small graph around any critical point.

Determine if the critical points are maxima, minima, or saddle points.

The function is (f(x,y)=x^2+2xy+4y^2+5x-6y ext<,>) for the region (-10le xle 10 ext<,>) and (-10le yle 10 ext<.>)

Set the partial derivatives equal to 0 and solve for (x) and (y ext<.>)

We can use either the method of substitution (solve for (x) or (y) in one equation and substitute into the other and solve), or method by elimination (multiply both equations by carefully chosen numbers and ass/subtract the equations from each other.)

We will demonstrate method of elimination:

Adding the two equations gives (6y-11=0 ext<.>) Hence (y= 11/6 ext<.>)

## Contents

Suppose that f is a function of more than one variable. For instance,

The graph of this function defines a surface in Euclidean space. To every point on this surface, there are an infinite number of tangent lines. Partial differentiation is the act of choosing one of these lines and finding its slope. Usually, the lines of most interest are those that are parallel to the x z -plane, and those that are parallel to the y z -plane (which result from holding either y or x constant, respectively).

### Basic definition Edit

The function f can be reinterpreted as a family of functions of one variable indexed by the other variables:

In other words, every value of y defines a function, denoted fy , which is a function of one variable x. [a] That is,

In this section the subscript notation fy denotes a function contingent on a fixed value of y, and not a partial derivative.

Once a value of y is chosen, say a, then f(x,y) determines a function fa which traces a curve x 2 + ax + a 2 on the x z -plane:

In this expression, a is a constant, not a variable, so fa is a function of only one real variable, that being x. Consequently, the definition of the derivative for a function of one variable applies:

The above procedure can be performed for any choice of a. Assembling the derivatives together into a function gives a function which describes the variation of f in the x direction:

This is the partial derivative of f with respect to x. Here is a rounded d called the partial derivative symbol. To distinguish it from the letter d, is sometimes pronounced "partial".

In general, the partial derivative of an n-ary function f(x1, …, xn) in the direction xi at the point (a1, …, an) is defined to be:

∂ f ∂ x i ( a 1 , … , a n ) = lim h → 0 f ( a 1 , … , a i + h , … , a n ) − f ( a 1 , … , a i , … , a n ) h >>(a_<1>,ldots ,a_)=lim _,ldots ,a_+h,ldots ,a_)-f(a_<1>,ldots ,a_,dots ,a_)>>> .

In the above difference quotient, all the variables except xi are held fixed. That choice of fixed values determines a function of one variable

In other words, the different choices of a index a family of one-variable functions just as in the example above. This expression also shows that the computation of partial derivatives reduces to the computation of one-variable derivatives.

This vector is called the gradient of f at a. If f is differentiable at every point in some domain, then the gradient is a vector-valued function ∇f which takes the point a to the vector ∇f(a). Consequently, the gradient produces a vector field.

### Formal definition Edit

∂ ∂ x i f ( a ) = lim h → 0 f ( a 1 , … , a i − 1 , a i + h , a i + 1 , … , a n ) − f ( a 1 , … , a i , … , a n ) h = lim h → 0 f ( a + h e i ) − f ( a ) h >>f(mathbf )&=lim _,ldots ,a_,a_+h,a_,ldots ,a_)-f(a_<1>,ldots ,a_,dots ,a_)>>&=lim _ )-f(mathbf )>>end>>

Even if all partial derivatives ∂f/∂xi(a) exist at a given point a, the function need not be continuous there. However, if all partial derivatives exist in a neighborhood of a and are continuous there, then f is totally differentiable in that neighborhood and the total derivative is continuous. In this case, it is said that f is a C 1 function. This can be used to generalize for vector valued functions, f : U → R m ^> , by carefully using a componentwise argument.

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