Calculus 1 Quiz: Higher Order Derivatives

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Higher Order DerivativesQuestion 1 of 20

If $e^y + y = x$ , what is $\frac{d^2y}{dx^2}$ in terms of $y$ ?

\frac{e^y}{(e^y+1)^2}

-\frac{e^y}{(e^y+1)^3}

-\frac{1}{(e^y+1)^2}

\frac{e^y}{(e^y+1)^3}

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Calculus 1 Quiz

Calculus 1 Quiz: Higher Order Derivatives

Practice Higher Order Derivatives in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Higher Order Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

If $e^y + y = x$ , what is $\frac{d^2y}{dx^2}$ in terms of $y$ ?

$\frac{e^y}{(e^y+1)^2}$
$-\frac{e^y}{(e^y+1)^3}$ (correct answer)
$-\frac{1}{(e^y+1)^2}$
$\frac{e^y}{(e^y+1)^3}$

Explanation: This problem tests implicit differentiation and finding second derivatives. When you have an equation like

e^y + y = x

where y is implicitly defined as a function of x, you need to differentiate twice to find the second derivative. First, find

\frac{dy}{dx}

by differentiating both sides with respect to x:

\frac{d}{dx}(e^y + y) = \frac{d}{dx}(x)

e^y \frac{dy}{dx} + \frac{dy}{dx} = 1

(e^y + 1)\frac{dy}{dx} = 1

\frac{dy}{dx} = \frac{1}{e^y + 1}

Now differentiate this result to find the second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{e^y + 1}\right)

Using the chain rule on this quotient:

\frac{d^2y}{dx^2} = -\frac{1}{(e^y + 1)^2} \cdot \frac{d}{dx}(e^y + 1) = -\frac{1}{(e^y + 1)^2} \cdot e^y \frac{dy}{dx}

Substituting

\frac{dy}{dx} = \frac{1}{e^y + 1}

\frac{d^2y}{dx^2} = -\frac{e^y}{(e^y + 1)^2} \cdot \frac{1}{e^y + 1} = -\frac{e^y}{(e^y + 1)^3}

This matches answer B. Answer A has the wrong sign and wrong power in the denominator. Answer C is missing the

e^y

factor in the numerator—this would be the result if you forgot to apply the chain rule when differentiating

e^y + 1

. Answer D has the correct expression but wrong sign. Remember: when finding second derivatives of implicit functions, carefully track your chain rule applications and substitute your first derivative back into the expression.

Question 2

Let $f(x) = x^2 \sin(x)$ . What is the fourth derivative, $f^{(4)}(x)$ ?

$(x^2 - 12)\sin(x) + 8x\cos(x)$
$(x^2 - 12)\sin(x) - 8x\cos(x)$ (correct answer)
$(12-x^2)\cos(x) - 8x\sin(x)$
$x^2\sin(x)$

Explanation: When you encounter a product of functions like

f(x) = x^2 \sin(x)

and need higher derivatives, you'll need to apply the product rule repeatedly. This requires careful tracking of patterns as derivatives of trigonometric and polynomial functions cycle through predictable forms. Let's work through each derivative systematically. Starting with

f(x) = x^2 \sin(x)

f'(x) = 2x\sin(x) + x^2\cos(x)

(product rule)

f''(x) = 2\sin(x) + 2x\cos(x) + 2x\cos(x) - x^2\sin(x) = 2\sin(x) + 4x\cos(x) - x^2\sin(x)

f'''(x) = 2\cos(x) + 4\cos(x) - 4x\sin(x) - 2x\sin(x) - x^2\cos(x) = 6\cos(x) - 6x\sin(x) - x^2\cos(x)

f^{(4)}(x) = -6\sin(x) - 6\sin(x) - 6x\cos(x) + 2x\cos(x) + x^2\sin(x) = (x^2 - 12)\sin(x) - 8x\cos(x)

This confirms answer B is correct. Answer A has the wrong sign on the cosine term—a common error when tracking signs through multiple applications of the product rule. Answer C mixes up the sine and cosine terms entirely, suggesting confusion about the derivatives of trigonometric functions. Answer D incorrectly suggests the fourth derivative equals the original function, which would only be true for

e^x

, not products involving trig functions. Study tip: When finding higher derivatives of products, work systematically and double-check your signs at each step. The derivatives of

\sin(x)

and

\cos(x)

follow a predictable cycle, but signs can easily get mixed up in complex expressions.

Question 3

For what value of $k$ does $y=e^{kx}$ satisfy the equation $y'' + y' - 6y = 0$ ?

$k=3$ or $k=-2$
$k=2$ or $k=-3$ (correct answer)
$k=6$ or $k=-1$
$k=1$ or $k=-6$

Explanation: When you encounter a differential equation with an exponential function guess, you're dealing with a characteristic equation problem. The key insight is that exponential functions maintain their form through differentiation, making them perfect candidates for linear differential equations. Let's substitute

y = e^{kx}

into the equation

y'' + y' - 6y = 0

. First, find the derivatives:

y' = ke^{kx}

and

y'' = k^2e^{kx}

. Substituting these into the differential equation gives us:

k^2e^{kx} + ke^{kx} - 6e^{kx} = 0

Factor out

e^{kx}

(which is never zero):

e^{kx}(k^2 + k - 6) = 0

This means

k^2 + k - 6 = 0

. This is called the characteristic equation. Factoring:

(k + 3)(k - 2) = 0

, so

k = -3

k = 2

. Looking at the wrong answers: Choice A gives

k = 3

k = -2

, which would satisfy

k^2 - k - 6 = 0

(wrong signs in the original equation). Choice C gives

k = 6

k = -1

, satisfying

k^2 - 5k - 6 = 0

(incorrect constant term). Choice D gives

k = 1

k = -6

, satisfying

k^2 + 5k - 6 = 0

(wrong middle coefficient). Study tip: Always set up the characteristic equation by substituting

y = e^{rx}

and its derivatives, then factor out the exponential term. The roots of the resulting polynomial give you the values that make the exponential function a solution.

Question 4

Given that $y' = x^2 + y^2$ , find $y''$ at the point $(1, 1)$ .

4
2
6 (correct answer)
8

Explanation: We are given

y' = x^2 + y^2

. To find

y''

, we differentiate this equation with respect to

x

y'' = 2x + 2y \cdot y'

. We need to evaluate this at the point

(1, 1)

. First, we find the value of

y'

(1, 1)

using the given equation:

y'(1,1) = 1^2 + 1^2 = 2

. Now we can find

y''

(1, 1)

by substituting

x=1

y=1

, and

y'=2

into the expression for

y''

y''(1,1) = 2(1) + 2(1)(2) = 2 + 4 = 6

Question 5

If $y = \arctan(x)$ , find an expression for $\frac{d^2y}{dx^2}$ .

$-\frac{2x}{1+x^2}$
$\frac{1}{(1+x^2)^2}$
$-\frac{2x}{(1+x^2)^2}$ (correct answer)
$-\frac{1}{(1+x^2)^2}$

Explanation: When finding the second derivative of inverse trigonometric functions, you'll need to apply the chain rule and quotient rule systematically. This type of problem tests your ability to differentiate composite functions multiple times. Starting with

y = \arctan(x)

, you first need the first derivative. The derivative of

\arctan(x)

\frac{dy}{dx} = \frac{1}{1+x^2}

. Now you need to differentiate this expression to find

\frac{d^2y}{dx^2}

. Since

\frac{dy}{dx} = (1+x^2)^{-1}

, use the chain rule:

\frac{d^2y}{dx^2} = -1 \cdot (1+x^2)^{-2} \cdot \frac{d}{dx}(1+x^2) = -(1+x^2)^{-2} \cdot 2x = -\frac{2x}{(1+x^2)^2}

. This confirms answer C is correct. Let's examine why the other choices are wrong. Choice A,

-\frac{2x}{1+x^2}

, represents what you'd get if you incorrectly applied the power rule without squaring the denominator—forgetting that differentiating

(1+x^2)^{-1}

requires the chain rule. Choice B,

\frac{1}{(1+x^2)^2}

, has the right denominator but wrong sign and numerator, suggesting confusion about the chain rule's negative contribution. Choice D,

-\frac{1}{(1+x^2)^2}

, gets the sign right but misses the

2x

factor from differentiating the inner function

1+x^2

. Remember: when differentiating inverse trig functions twice, the second derivative will typically involve the derivative of the "inside" function (here,

2x

) appearing in the numerator due to the chain rule.

Question 6

A function $y=f(x)$ satisfies the differential equation $y'' - 2y' + y = 0$ . If $f(0)=1$ and $f'(0)=3$ , what is the value of $f'''(0)$ ?

7 (correct answer)
5
9
3

Explanation: First, we use the given differential equation to find

y''(0)

. At

x=0

y''(0) - 2y'(0) + y(0) = 0

. Substituting the given values:

y''(0) - 2(3) + 1 = 0

, which gives

y''(0) = 5

. Next, we differentiate the entire differential equation with respect to

x

to find a relationship for

y'''

y''' - 2y'' + y' = 0

. This can be rearranged to

y''' = 2y'' - y'

. Now, we can find

y'''(0)

by substituting the values we know:

y'''(0) = 2y''(0) - y'(0) = 2(5) - 3 = 10 - 3 = 7

Question 7

Let $g(x) = x \ln(x)$ . What is the value of $g^{(4)}(1)$ , the fourth derivative of $g(x)$ evaluated at $x=1$ ?

$-1$
$1$
$2$ (correct answer)
$-2$

Explanation: When you encounter a problem asking for a higher-order derivative, you need to systematically compute each derivative step by step until you reach the desired order. Starting with

g(x) = x \ln(x)

, let's find each derivative using the product rule where needed: First derivative:

g'(x) = \frac{d}{dx}[x \ln(x)] = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1

Second derivative:

g''(x) = \frac{d}{dx}[\ln(x) + 1] = \frac{1}{x} + 0 = \frac{1}{x}

Third derivative:

g'''(x) = \frac{d}{dx}\left[\frac{1}{x}\right] = \frac{d}{dx}[x^{-1}] = -x^{-2} = -\frac{1}{x^2}

Fourth derivative:

g^{(4)}(x) = \frac{d}{dx}\left[-\frac{1}{x^2}\right] = \frac{d}{dx}[-x^{-2}] = -(-2)x^{-3} = \frac{2}{x^3}

Now evaluating at

x = 1

g^{(4)}(1) = \frac{2}{1^3} = 2

Answer choice A)

-1

might come from confusing signs or stopping at the third derivative and evaluating incorrectly. Answer choice B) 1 could result from missing the coefficient 2 in the final derivative. Answer choice D)

-2

represents getting the magnitude right but making a sign error when differentiating

-\frac{1}{x^2}

. Study tip: For higher-order derivatives, work methodically one step at a time and double-check your signs. Pattern recognition helps too—derivatives of

x^n

follow predictable factorial-like patterns.

Question 8

Let $f(x) = \sqrt{2x+1}$ . Find the value of $f''(4)$ .

$\frac{1}{9}$
$\frac{1}{3}$
$-\frac{1}{54}$
$-\frac{1}{27}$ (correct answer)

Explanation: This question tests your ability to find second derivatives of functions involving radicals, which requires careful application of the chain rule and power rule multiple times. To find

f''(4)

where

f(x) = \sqrt{2x+1}

, you'll need to compute the first and second derivatives step by step. Start by rewriting the function using fractional exponents:

f(x) = (2x+1)^{1/2}

. For the first derivative, apply the chain rule:

f'(x) = \frac{1}{2}(2x+1)^{-1/2} \cdot 2 = (2x+1)^{-1/2}

. For the second derivative, differentiate again using the chain rule:

f''(x) = -\frac{1}{2}(2x+1)^{-3/2} \cdot 2 = -(2x+1)^{-3/2}

. Now substitute

x = 4

f''(4) = -(2 \cdot 4 + 1)^{-3/2} = -(9)^{-3/2} = -\frac{1}{9^{3/2}} = -\frac{1}{27}

. Answer A gives

\frac{1}{9}

, which might result from incorrectly computing only

f'(4)

and making a sign error. Answer B gives

\frac{1}{3}

, which could come from confusing the exponent in the derivative. Answer C gives

-\frac{1}{54}

, which might result from an arithmetic error when simplifying

9^{3/2}

or incorrectly handling the coefficient from the chain rule. The correct answer is D:

-\frac{1}{27}

. Remember that when differentiating radical functions, converting to fractional exponents first makes the process cleaner, and always double-check your chain rule applications—the coefficients multiply at each step.

Question 9

If $f(x) = x^4 - 2x^3 + 5x - 1$ , find the value of $f'''(2)$ .

24
36 (correct answer)
12
48

Explanation: This question tests your ability to find higher-order derivatives, specifically the third derivative of a polynomial function. When you see a question asking for

f'''(x)

or the third derivative, you'll need to differentiate the original function three times, then evaluate at the given point. Let's work through this systematically. Starting with

f(x) = x^4 - 2x^3 + 5x - 1

, we find each derivative: First derivative:

f'(x) = 4x^3 - 6x^2 + 5

Second derivative:

f''(x) = 12x^2 - 12x

Third derivative:

f'''(x) = 24x - 12

Now evaluate at

x = 2

f'''(2) = 24(2) - 12 = 48 - 12 = 36

Looking at the wrong answers: Choice A (24) likely comes from forgetting to subtract 12 and only calculating

24 \times 2 = 48

, then making an arithmetic error. Choice C (12) might result from incorrectly finding the third derivative as just the constant term

-12

and taking its absolute value, or from errors in the differentiation process. Choice D (48) is what you'd get if you calculated

24 \times 2

but forgot to subtract the 12. Study tip: When finding higher-order derivatives of polynomials, work step-by-step and don't rush. The power rule makes this mechanical: bring down the exponent and reduce it by 1. Double-check each derivative before moving to the next, and always perform the final substitution carefully—arithmetic errors are common at this last step.

Question 10

Let $f(x) = e^{x^2}$ . What is the value of $f''(2)$ ?

$18e^4$ (correct answer)
$16e^4$
$8e^4$
$4e^4$

Explanation: First, find the first derivative using the chain rule:

f'(x) = e^{x^2} \cdot 2x = 2xe^{x^2}

. Next, find the second derivative using the product rule and chain rule:

f''(x) = (2)(e^{x^2}) + (2x)(e^{x^2} \cdot 2x) = 2e^{x^2} + 4x^2e^{x^2} = (2 + 4x^2)e^{x^2}

. Finally, evaluate at

x=2

f''(2) = (2 + 4(2^2))e^{2^2} = (2 + 16)e^4 = 18e^4

Question 11

If $g(x) = f^{-1}(x)$ and $f(2)=5$ , $f'(2)=3$ , and $f''(2)=-1$ , what is the value of $g''(5)$ ?

$-\frac{1}{9}$
$-\frac{1}{27}$
$\frac{1}{9}$
$\frac{1}{27}$ (correct answer)

Explanation: When you encounter inverse functions and their derivatives, remember that finding the second derivative of an inverse function requires applying the derivative formula for inverse functions twice. Since

g(x) = f^{-1}(x)

, we need

g''(5)

. First, use the inverse function derivative formula:

g'(x) = \frac{1}{f'(g(x))}

. Since

f(2) = 5

, we know

g(5) = 2

, so

g'(5) = \frac{1}{f'(2)} = \frac{1}{3}

. For the second derivative, differentiate

g'(x) = \frac{1}{f'(g(x))}

using the chain rule:

g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2}

. Substituting

x = 5

g''(5) = -\frac{f''(g(5)) \cdot g'(5)}{[f'(g(5))]^2} = -\frac{f''(2) \cdot \frac{1}{3}}{[f'(2)]^2} = -\frac{(-1) \cdot \frac{1}{3}}{3^2} = -\frac{-\frac{1}{3}}{9} = \frac{1}{27}

. Choice A gives

-\frac{1}{9}

, which you'd get if you forgot the negative sign from

f''(2) = -1

and miscalculated the denominator. Choice B gives

-\frac{1}{27}

, the result if you incorrectly kept a negative sign. Choice C gives

\frac{1}{9}

, which occurs from using

[f'(2)]^1

instead of

[f'(2)]^2

in the denominator. The correct answer is D:

\frac{1}{27}

. Key strategy: Always remember the second derivative formula for inverse functions involves both

f'

and

f''

, and practice tracking negative signs carefully through the chain rule calculations.

Question 12

Let $f(x) = \sin(2x)$ . Find $f^{(15)}(x)$ , the 15th derivative of $f(x)$ .

$-2^{15}\cos(2x)$ (correct answer)
$2^{15}\cos(2x)$
$-2^{15}\sin(2x)$
$2^{15}\sin(2x)$

Explanation: The derivatives of

\sin(2x)

follow a cycle of four:

f'(x) = 2\cos(2x)

f''(x) = -4\sin(2x)

f'''(x) = -8\cos(2x)

f^{(4)}(x) = 16\sin(2x)

. The general pattern involves a factor of

2^n

and a repeating trigonometric part. To find the 15th derivative, we find the remainder of

15

divided by

4

, which is

3

(since

15 = 4 \cdot 3 + 3

). Therefore, the 15th derivative will follow the pattern of the 3rd derivative. The 3rd derivative is

-2^3\cos(2x)

, so the 15th derivative is

f^{(15)}(x) = -2^{15}\cos(2x)

Question 13

Let $f(x) = \frac{x}{x+1}$ . Find $f^{(10)}(1)$ .

$\frac{10!}{2^{11}}$
$-\frac{10!}{2^{11}}$ (correct answer)
$\frac{9!}{2^{10}}$
$-\frac{9!}{2^{10}}$

Explanation: First, rewrite

f(x) = \frac{x+1-1}{x+1} = 1 - \frac{1}{x+1} = 1 - (x+1)^{-1}

. The derivative of the constant 1 is zero, so we only need to find the 10th derivative of

-(x+1)^{-1}

. Let

g(x) = -(x+1)^{-1}

g'(x) = (x+1)^{-2}

g''(x) = -2(x+1)^{-3}

g'''(x) = 6(x+1)^{-4}

. The pattern is

g^{(n)}(x) = (-1)^{n+1} n! (x+1)^{-(n+1)}

. For

n=10

, we have

f^{(10)}(x) = g^{(10)}(x) = (-1)^{11} 10! (x+1)^{-11} = -\frac{10!}{(x+1)^{11}}

. Evaluating at

x=1

gives

f^{(10)}(1) = -\frac{10!}{(1+1)^{11}} = -\frac{10!}{2^{11}}

Question 14

Let $f(x) = \ln(1-x)$ . Find the value of $f^{(5)}(0)$ , the fifth derivative of $f$ evaluated at $x=0$ .

120
24
-120
-24 (correct answer)

Explanation: Finding higher-order derivatives of logarithmic functions requires recognizing patterns and applying systematic differentiation. When you encounter

f(x) = \ln(1-x)

, the key is to find the derivatives step by step and look for a pattern. Let's compute the derivatives systematically:

$f(x) = \ln(1-x)$
$f'(x) = \frac{-1}{1-x} = -(1-x)^{-1}$
$f''(x) = -(1)(1-x)^{-2} \cdot (-1) = -(1-x)^{-2}$
$f'''(x) = -(-2)(1-x)^{-3} \cdot (-1) = -2(1-x)^{-3}$
$f^{(4)}(x) = -2(-3)(1-x)^{-4} \cdot (-1) = -6(1-x)^{-4}$
$f^{(5)}(x) = -6(-4)(1-x)^{-5} \cdot (-1) = -24(1-x)^{-5}$

Evaluating at

x = 0

f^{(5)}(0) = -24(1-0)^{-5} = -24

, confirming answer (D). The wrong answers represent common calculation errors: (A) 120 likely comes from forgetting the negative sign and miscalculating the factorial pattern. (B) 24 results from dropping the negative sign entirely. (C) -120 suggests correctly identifying the negative sign but incorrectly computing

4! = 24

5! = 120

. Study tip: For derivatives of

\ln(1-x)

, the

n

-th derivative follows the pattern

f^{(n)}(x) = -(n-1)!(1-x)^{-n}

. Always track your negative signs carefully and remember that the factorial is

(n-1)!

, not

n!

Question 15

The position of a particle moving along a line is given by $s(t) = t^2 e^{-t}$ for $t \ge 0$ . The jerk of the particle is defined as the third derivative of position, $s'''(t)$ . What is the smallest time $t>0$ at which the particle's jerk is zero?

$1 + \sqrt{3}$
$2 - \sqrt{2}$
$2$
$3 - \sqrt{3}$ (correct answer)

Explanation: When you encounter problems involving jerk (the third derivative of position), you need to systematically take derivatives and then solve for when that third derivative equals zero. Starting with

s(t) = t^2 e^{-t}

, you'll need to find

s'''(t)

. For the first derivative, use the product rule:

s'(t) = 2t e^{-t} + t^2(-e^{-t}) = e^{-t}(2t - t^2)

For the second derivative, apply the product rule again:

s''(t) = -e^{-t}(2t - t^2) + e^{-t}(2 - 2t) = e^{-t}(t^2 - 4t + 2)

For the third derivative:

s'''(t) = -e^{-t}(t^2 - 4t + 2) + e^{-t}(2t - 4) = e^{-t}(-t^2 + 6t - 6)

Setting

s'''(t) = 0

: Since

e^{-t} \neq 0

, you need

-t^2 + 6t - 6 = 0

, or

t^2 - 6t + 6 = 0

. Using the quadratic formula:

t = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3}

Since you want the smallest positive time, choose

t = 3 - \sqrt{3}

(answer D). Answer A (

1 + \sqrt{3} \approx 2.73

) and answer C (

2

) are smaller values that don't satisfy our equation. Answer B (

2 - \sqrt{2} \approx 0.59

) is the smallest numerically but also doesn't solve our quadratic equation. Remember: with higher-order derivatives of products involving exponentials, always factor out the exponential term before setting equal to zero—it simplifies the algebra significantly.

Question 16

If $h(x) = x^2 \cos(x)$ , find $h'''(x)$ .

$(x^2 - 6)\sin(x) - 6x\cos(x)$ (correct answer)
$(6-x^2)\sin(x) + 6x\cos(x)$
$x^2\sin(x) - 6x\cos(x) - 6\sin(x)$
$-x^2\cos(x) - 4x\sin(x) + 2\cos(x)$

Explanation: We differentiate three times:

h'(x) = 2x\cos(x) - x^2\sin(x)

. For the second derivative:

h''(x) = (2\cos(x) - 2x\sin(x)) - (2x\sin(x) + x^2\cos(x)) = 2\cos(x) - 4x\sin(x) - x^2\cos(x)

. For the third derivative:

h'''(x) = (-2\sin(x)) - (4\sin(x) + 4x\cos(x)) - (2x\cos(x) - x^2\sin(x)) = -2\sin(x) - 4\sin(x) - 4x\cos(x) - 2x\cos(x) + x^2\sin(x) = (x^2 - 6)\sin(x) - 6x\cos(x)

Question 17

If $x^2 + y^2 = 25$ , what is the value of $\frac{d^2y}{dx^2}$ at the point $(3, -4)$ ?

$\frac{25}{64}$ (correct answer)
$-\frac{25}{64}$
$\frac{7}{16}$
$\frac{3}{4}$

Explanation: Differentiating the equation

x^2 + y^2 = 25

implicitly with respect to

x

gives

2x + 2y \frac{dy}{dx} = 0

, which solves to

\frac{dy}{dx} = -\frac{x}{y}

. To find the second derivative, we differentiate this expression using the quotient rule:

\frac{d^2y}{dx^2} = -\frac{(1)y - x(\frac{dy}{dx})}{y^2} = -\frac{y - x(-\frac{x}{y})}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}

. Since we know from the original equation that

x^2 + y^2 = 25

, we can substitute this in:

\frac{d^2y}{dx^2} = -\frac{25}{y^3}

. At the point

(3, -4)

, we have

y = -4

, so

\frac{d^2y}{dx^2} = -\frac{25}{(-4)^3} = -\frac{25}{-64} = \frac{25}{64}

Question 18

Let $f(x) = (x+2)^5$ . The fourth derivative $f^{(4)}(x)$ can be written as $k(x+2)^n$ . What is the value of $k+n$ ?

60
122
61
121 (correct answer)

Explanation: When you encounter a function like

f(x) = (x+2)^5

and need to find higher-order derivatives, you're dealing with the power rule applied repeatedly to a composite function. Let's find the fourth derivative step by step. Using the chain rule with the power rule:

f'(x) = 5(x+2)^4 \cdot 1 = 5(x+2)^4

f''(x) = 5 \cdot 4(x+2)^3 = 20(x+2)^3

f'''(x) = 20 \cdot 3(x+2)^2 = 60(x+2)^2

f^{(4)}(x) = 60 \cdot 2(x+2)^1 = 120(x+2)^1

f^{(4)}(x) = 120(x+2)^1

, which means

k = 120

and

n = 1

. Therefore,

k + n = 120 + 1 = 121

. Choice A (60) represents just the coefficient from the third derivative, stopping one step too early. Choice B (122) likely comes from miscounting derivatives or making an arithmetic error with

120 + 2 = 122

. Choice C (61) might result from confusing the pattern and thinking

k = 60

(the third derivative coefficient) plus

n = 1

. The correct answer is D (121). Remember this pattern: when differentiating

(x+a)^n

, each derivative multiplies by the current exponent and reduces the exponent by 1. For the

r

-th derivative of

(x+a)^n

, the coefficient follows the falling factorial:

n(n-1)(n-2)...(n-r+1)

. This systematic approach prevents calculation errors in higher-order derivatives.

Question 19

For the curve defined by the equation $x^3 + y^3 = 9$ , what is the value of $\frac{d^2y}{dx^2}$ at the point $(1, 2)$ ?

$-\frac{9}{16}$ (correct answer)
$-\frac{1}{2}$
$-\frac{7}{16}$
$\frac{9}{16}$

Explanation: First, we find the first derivative using implicit differentiation:

3x^2 + 3y^2 \frac{dy}{dx} = 0

, which simplifies to

\frac{dy}{dx} = -\frac{x^2}{y^2}

. At the point

(1, 2)

, we have

\frac{dy}{dx} = -\frac{1^2}{2^2} = -\frac{1}{4}

. Next, we differentiate

\frac{dy}{dx}

with respect to

x

using the quotient rule to find the second derivative:

\frac{d^2y}{dx^2} = -\frac{(2x)(y^2) - (x^2)(2y \frac{dy}{dx})}{ (y^2)^2 } = -\frac{2xy^2 - 2x^2y \frac{dy}{dx}}{y^4}

Now, we substitute the values from the point

(1, 2)

and the value of

\frac{dy}{dx} = -\frac{1}{4}

\frac{d^2y}{dx^2} = -\frac{2(1)(2^2) - 2(1^2)(2)(-\frac{1}{4})}{2^4} = -\frac{8 - 4(-\frac{1}{4})}{16} = -\frac{8 - (-1)}{16} = -\frac{8+1}{16} = -\frac{9}{16}

Question 20

The $n$ -th derivative of $f(x) = x e^x$ is given by $f^{(n)}(x) = (x+n)e^x$ . Using this fact, what is $f^{(100)}(0)$ ?

0
100!
1
100 (correct answer)

Explanation: When you encounter a problem that gives you a formula for the n-th derivative, you're being tested on your ability to apply that formula directly and evaluate derivatives at specific points. Given that

f^{(n)}(x) = (x+n)e^x

, you can find the 100th derivative by substituting

n = 100

f^{(100)}(x) = (x+100)e^x

. To evaluate this at

x = 0

, substitute:

f^{(100)}(0) = (0+100)e^0 = 100 \cdot 1 = 100

. Let's examine why the other answers are incorrect. Choice (A) gives 0, which would only be true if the coefficient of

e^x

were zero - but

(0+100) = 100 \neq 0

. Choice (B) suggests 100!, which might seem reasonable since factorials often appear in derivative formulas, but this specific formula

(x+n)e^x

doesn't involve factorials when evaluated at zero. Choice (C) gives 1, which you might get if you forgot about the

(x+n)

term and only considered

e^0 = 1

, but you can't ignore the coefficient. The key insight is recognizing that when

x = 0

, the exponential term

e^0

always equals 1, so the entire expression simplifies to just the coefficient

(0+n) = n

. This makes the calculation straightforward: the 100th derivative of

xe^x

evaluated at zero is simply 100. Study tip: When evaluating derivatives at

x = 0

, remember that

e^0 = 1

, which often simplifies complex exponential expressions to just their coefficients.