IB Mathematics: Analysis and Approaches Quiz: Differentiation Rules

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Differentiation RulesQuestion 1 of 20

Find the derivative of $f(x) = \cos(\frac{1}{x})$ .

-\sin(\frac{1}{x})

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

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

\sin(\frac{1}{x})

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Differentiation Rules

Practice Differentiation Rules in IB Mathematics: Analysis and Approaches 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 Differentiation Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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

Find the derivative of $f(x) = \cos(\frac{1}{x})$ .

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

Explanation: We must use the chain rule. Let the outer function be

g(u) = \cos(u)

and the inner function be

u(x) = \frac{1}{x} = x^{-1}

. The derivatives are:

g'(u) = -\sin(u)

u'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}

By the chain rule,

f'(x) = g'(u(x)) \cdot u'(x)

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

The two negative signs cancel out:

f'(x) = \frac{1}{x^2}\sin(\frac{1}{x})

A: $-\sin(\frac{1}{x})$ is incorrect. This happens if the derivative of the inner function $\frac{1}{x}$ is forgotten, or assumed to be 1.
B: $\frac{1}{x^2}\sin(\frac{1}{x})$ is correct.
C: $-\frac{1}{x^2}\sin(\frac{1}{x})$ is incorrect. This results from making only one sign error, either in differentiating $\cos(u)$ or in differentiating $\frac{1}{x}$ , but not both.
D: $\sin(\frac{1}{x})$ is incorrect. This results from a sign error in the derivative of $\cos(u)$ and also forgetting to differentiate the inner function.

Question 2

Let $f(x) = \sqrt{\sin(x) + \cos(x)}$ . Find the value of $f'(\frac{\pi}{4})$ .

0 (correct answer)
$\frac{1}{2}$
$\frac{\sqrt{2}}{2}$
$2^{1/4}$

Explanation: First, write the function using a power:

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

. Now, use the chain rule.

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

f'(x) = \frac{\cos x - \sin x}{2\sqrt{\sin x + \cos x}}

Now, substitute

x = \frac{\pi}{4}

. At this value,

\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}

and

\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}

f'(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{2\sqrt{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}}

f'(\frac{\pi}{4}) = \frac{0}{2\sqrt{\sqrt{2}}} = 0

A: 0 is correct.
B: $\frac{1}{2}$ is incorrect. This might be a guess or result from a calculation error.
C: $\frac{\sqrt{2}}{2}$ is incorrect. This is the value of $\cos(\frac{\pi}{4})$ and $\sin(\frac{\pi}{4})$ .
D: $2^{1/4}$ is incorrect. This is the value of the original function $f(\frac{\pi}{4}) = \sqrt{\sqrt{2}} = (2^{1/2})^{1/2} = 2^{1/4}$ , not its derivative.

Question 3

Find the gradient of the tangent to the curve $y = (x^2 + 1)\sin(x)$ at the point where $x = \pi$ .

$-(\pi^2+1)$ (correct answer)
$-2\pi$
0
$\pi^2+1$

Explanation: First, find the derivative

\frac{dy}{dx}

using the product rule,

(uv)' = u'v + uv'

, with

u = x^2+1

and

v = \sin(x)

u' = 2x

v' = \cos(x)

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

Now, evaluate the derivative at

x = \pi

. We use the values

\sin(\pi) = 0

and

\cos(\pi) = -1

. Gradient at

x=\pi

is:

\frac{dy}{dx}\Big|_{x=\pi} = (2\pi)(\sin(\pi)) + (\pi^2+1)(\cos(\pi))

= (2\pi)(0) + (\pi^2+1)(-1)

= -(\pi^2+1)

A: $-(\pi^2+1)$ is correct.
B: $-2\pi$ is incorrect. This comes from an incorrect application of the product rule, for example $u'v' = (2x)(\cos x)$ , which evaluates to $2\pi(-1) = -2\pi$ at $x=\pi$ .
C: 0 is incorrect. This is the value of the function $y$ at $x=\pi$ , not its gradient.
D: $\pi^2+1$ is incorrect. This results from a sign error, likely by evaluating $\cos(\pi)$ as $+1$ instead of $-1$ .

Question 4

The function $f$ is given by $f(x) = \frac{e^x}{e^x + 1}$ . Find $f'(x)$ .

$\frac{-e^x}{(e^x+1)^2}$
1
0
$\frac{e^x}{(e^x+1)^2}$ (correct answer)

Explanation: Use the quotient rule,

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

, with

u = e^x

and

v = e^x + 1

. First, find the derivatives of

u

and

v

u' = e^x

v' = e^x

Apply the quotient rule:

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

Simplify the numerator:

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

f'(x) = \frac{e^x}{(e^x+1)^2}

A: $\frac{-e^x}{(e^x+1)^2}$ is incorrect. This results from reversing the terms in the numerator of the quotient rule to $uv' - u'v$ .
B: 1 is incorrect. This might arise from incorrectly differentiating the numerator and denominator separately and then dividing: $\frac{e^x}{e^x} = 1$ .
C: 0 is incorrect. This might arise from an algebraic error in the numerator, such as $e^x(e^x+1) - e^x(e^x+1) = 0$ .
D: $\frac{e^x}{(e^x+1)^2}$ is correct.

Question 5

Find the x-coordinates of the stationary points of the function $f(x) = x^3 e^{-x}$ .

x = 3 only
x = 0 only
x = 0 and x = 3 (correct answer)
x = 0 and x = -3

Explanation: To find the stationary points, we need to find the derivative

f'(x)

and set it to zero. Use the product rule with

u = x^3

and

v = e^{-x}

u' = 3x^2

v' = -e^{-x}

(using the chain rule)

f'(x) = u'v + uv' = (3x^2)(e^{-x}) + (x^3)(-e^{-x})

Now, factor the expression:

f'(x) = e^{-x}(3x^2 - x^3) = x^2e^{-x}(3-x)

Set

f'(x) = 0

x^2e^{-x}(3-x) = 0

Since

e^{-x}

is never zero, we have two possibilities for the product to be zero:

$x^2 = 0 \implies x = 0$
$3-x = 0 \implies x = 3$

So, the x-coordinates of the stationary points are 0 and 3.

A: x = 3 only is incorrect. This ignores the solution from the $x^2$ factor.
B: x = 0 only is incorrect. This ignores the solution from the $(3-x)$ factor.
C: x = 0 and x = 3 is correct.
D: x = 0 and x = -3 is incorrect. This comes from a sign error when solving $3-x=0$ .

Question 6

The derivative of $f(x) = (ax+b)\cos(x)$ is $f'(x) = 2\cos(x) - (2x+1)\sin(x)$ . Find the values of $a$ and $b$ .

a = -2, b = -1
a = 2, b = -1
a = 1, b = 2
a = 2, b = 1 (correct answer)

Explanation: First, find the derivative of

f(x) = (ax+b)\cos(x)

using the product rule. Let

u = ax+b

and

v = \cos(x)

u' = a

v' = -\sin(x)

f'(x) = u'v + uv' = (a)(\cos(x)) + (ax+b)(-\sin(x))

f'(x) = a\cos(x) - (ax+b)\sin(x)

We are given that

f'(x) = 2\cos(x) - (2x+1)\sin(x)

. We can equate the two expressions for

f'(x)

a\cos(x) - (ax+b)\sin(x) = 2\cos(x) - (2x+1)\sin(x)

By comparing the coefficients of the

\cos(x)

terms, we get:

a = 2

By comparing the coefficients of the

\sin(x)

terms (the expressions inside the parentheses), we get:

ax+b = 2x+1

Substitute

a=2

into this equation:

2x+b = 2x+1

This implies that

b=1

. So,

a=2

and

b=1

A: a = -2, b = -1 is incorrect, resulting from sign errors.
B: a = 2, b = -1 is incorrect, resulting from a sign error for b.
C: a = 1, b = 2 is incorrect, mixing up the values of a and b.
D: a = 2, b = 1 is correct.

Question 7

Let $h(x) = x \cos(2x^2)$ . Find $h'(x)$ .

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

Explanation: We use the product rule,

(uv)' = u'v + uv'

, with

u = x

and

v = \cos(2x^2)

u' = 1

To find

v'

, we use the chain rule. The derivative of

\cos(w)

-\sin(w) \cdot w'

. Here,

w = 2x^2

, so

w' = 4x

v' = -\sin(2x^2) \cdot 4x = -4x\sin(2x^2)

Now, apply the product rule:

h'(x) = (1)(\cos(2x^2)) + (x)(-4x\sin(2x^2))

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

A: $\cos(2x^2) - 4x^2\sin(2x^2)$ is correct.
B: $\cos(2x^2) + 4x^2\sin(2x^2)$ is incorrect. This is a sign error from the derivative of $\cos(w)$ , treating it as $\sin(w)$ .
C: $\cos(2x^2) - 2x^2\sin(2x^2)$ is incorrect. This results from an error in the chain rule for $v$ , differentiating $2x^2$ as $2x$ instead of $4x$ .
D: $-4x \sin(2x^2)$ is incorrect. This is only the derivative of the second part, $\cos(2x^2)$ , and ignores the product rule.

Question 8

Let $f(x) = x \ln x$ for $x>0$ . Find $f''(x)$ .

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

Explanation: To find the second derivative,

f''(x)

, we must first find the first derivative,

f'(x)

. Using the product rule on

f(x) = x \ln x

with

u=x

and

v=\ln x

u'=1

v'=1/x

f'(x) = u'v + uv' = (1)(\ln x) + (x)(\frac{1}{x}) = \ln x + 1

Now, differentiate

f'(x)

to find

f''(x)

f''(x) = \frac{d}{dx}(\ln x + 1) = \frac{1}{x} + 0 = \frac{1}{x}

A: $-\frac{1}{x^2}$ is incorrect. This would be the second derivative if the product rule was ignored and $f'(x)$ was mistakenly taken as just $1/x$ .
B: $\frac{1}{x}$ is correct.
C: 1 is incorrect. This may result from incorrectly differentiating $\ln x$ as 1.
D: $\ln x + 1$ is incorrect. This is the first derivative, $f'(x)$ , not the second.

Question 9

Find the derivative of $f(x) = \cos(\frac{1}{x})$ .

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

Explanation: We must use the chain rule. Let the outer function be

g(u) = \cos(u)

and the inner function be

u(x) = \frac{1}{x} = x^{-1}

. The derivatives are:

g'(u) = -\sin(u)

u'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}

By the chain rule,

f'(x) = g'(u(x)) \cdot u'(x)

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

The two negative signs cancel out:

f'(x) = \frac{1}{x^2}\sin(\frac{1}{x})

A: $-\sin(\frac{1}{x})$ is incorrect. This happens if the derivative of the inner function $\frac{1}{x}$ is forgotten, or assumed to be 1.
B: $\frac{1}{x^2}\sin(\frac{1}{x})$ is correct.
C: $-\frac{1}{x^2}\sin(\frac{1}{x})$ is incorrect. This results from making only one sign error, either in differentiating $\cos(u)$ or in differentiating $\frac{1}{x}$ , but not both.
D: $\sin(\frac{1}{x})$ is incorrect. This results from a sign error in the derivative of $\cos(u)$ and also forgetting to differentiate the inner function.

Question 10

Let $f(x) = x^2 e^{3x}$ . Find $f'(x)$ .

$f'(x) = 6x e^{3x}$
$f'(x) = e^{3x}(2x + x^2)$
$f'(x) = e^{3x}(2x + 3x^2)$ (correct answer)
$f'(x) = e^{3x}(2 + 3x)$

Explanation: To find the derivative of

f(x) = x^2 e^{3x}

, we must use the product rule,

(uv)' = u'v + uv'

, where

u = x^2

and

v = e^{3x}

. First, find the derivatives of

u

and

v

u' = 2x

To find

v'

, we need the chain rule. The derivative of

e^{3x}

e^{3x} \cdot 3 = 3e^{3x}

. Now, apply the product rule:

f'(x) = (2x)(e^{3x}) + (x^2)(3e^{3x})

Finally, factor out the common term

xe^{3x}

for the simplified form, but the answer choices have

e^{3x}

factored out.

f'(x) = e^{3x}(2x + 3x^2)

A: $f'(x) = 6x e^{3x}$ is incorrect. This results from incorrectly multiplying the derivatives of the two parts, i.e., $(2x)(3e^{3x})$ .
B: $f'(x) = e^{3x}(2x + x^2)$ is incorrect. This happens if the chain rule is forgotten when differentiating $e^{3x}$ , treating its derivative as just $e^{3x}$ .
C: $f'(x) = e^{3x}(2x + 3x^2)$ is correct.
D: $f'(x) = e^{3x}(2 + 3x)$ is incorrect. This would arise from incorrectly factoring $x$ out of $2x + 3x^2$ .

Question 11

Given $y = \frac{\ln x}{x^3}$ for $x > 0$ , find $\frac{dy}{dx}$ .

$\frac{1 - 3\ln x}{x^4}$ (correct answer)
$\frac{3\ln x - 1}{x^4}$
$\frac{1 - 3x^2 \ln x}{x^6}$
$\frac{1 - 3\ln x}{x^3}$

Explanation: We use the quotient rule,

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

, with

u = \ln x

and

v = x^3

. First, find the derivatives of

u

and

v

u' = \frac{1}{x}

v' = 3x^2

Apply the quotient rule:

\frac{dy}{dx} = \frac{(\frac{1}{x})(x^3) - (\ln x)(3x^2)}{(x^3)^2}

Now, simplify the expression:

\frac{dy}{dx} = \frac{x^2 - 3x^2 \ln x}{x^6}

Factor out

x^2

from the numerator:

\frac{dy}{dx} = \frac{x^2(1 - 3\ln x)}{x^6}

Cancel the

x^2

term:

\frac{dy}{dx} = \frac{1 - 3\ln x}{x^4}

A: $\frac{1 - 3\ln x}{x^4}$ is correct.
B: $\frac{3\ln x - 1}{x^4}$ is incorrect. This results from reversing the terms in the numerator of the quotient rule, calculating $\frac{uv' - u'v}{v^2}$ .
C: $\frac{1 - 3x^2 \ln x}{x^6}$ is incorrect. This is an error in simplifying the term $\frac{1}{x} \cdot x^3$ in the numerator.
D: $\frac{1 - 3\ln x}{x^3}$ is incorrect. This results from forgetting to square the denominator $v$ in the quotient rule.

Question 12

Let $f(x) = \sin(\ln x)$ for $x > 0$ . Find the value of $f'(e^{\pi/2})$ .

$-e^{-\pi/2}$
0 (correct answer)
1
$e^{-\pi/2}$

Explanation: First, find the derivative

f'(x)

using the chain rule. Let

u = \ln x

. Then

f(x) = \sin(u)

f'(x) = \cos(u) \cdot u' = \cos(\ln x) \cdot \frac{1}{x} = \frac{\cos(\ln x)}{x}

Now, evaluate this derivative at

x = e^{\pi/2}

f'(e^{\pi/2}) = \frac{\cos(\ln(e^{\pi/2}))}{e^{\pi/2}}

Since

\ln(e^{\pi/2}) = \pi/2

f'(e^{\pi/2}) = \frac{\cos(\pi/2)}{e^{\pi/2}}

We know that

\cos(\pi/2) = 0

, so:

f'(e^{\pi/2}) = \frac{0}{e^{\pi/2}} = 0

A: $-e^{-\pi/2}$ is incorrect. This could arise from miscalculation, perhaps at a different point like $x = e^{\pi}$ .
B: 0 is correct.
C: 1 is incorrect. This is the value of the function $f(e^{\pi/2}) = \sin(\ln(e^{\pi/2})) = \sin(\pi/2) = 1$ , not its derivative.
D: $e^{-\pi/2}$ is incorrect. This is the value of $\frac{1}{x}$ at $x = e^{\pi/2}$ if the $\cos(\ln x)$ term in the numerator was mistakenly evaluated as 1.

Question 13

Let $y = \ln(\sin(e^x))$ . Which of the following is $\frac{dy}{dx}$ ?

$\cot(e^x)$
$e^x \tan(e^x)$
$\frac{e^x}{\sin(e^x)}$
$e^x \cot(e^x)$ (correct answer)

Explanation: This is a nested chain rule problem. Let's differentiate from the outermost function inwards.

The derivative of $\ln(u)$ is $\frac{1}{u}$ . Here $u = \sin(e^x)$ .
The derivative of $\sin(v)$ is $\cos(v)$ . Here $v = e^x$ .
The derivative of $e^x$ is $e^x$ .

Applying the chain rule, we multiply these derivatives together:

\frac{dy}{dx} = \frac{1}{\sin(e^x)} \cdot \cos(e^x) \cdot e^x

Simplifying this expression:

\frac{dy}{dx} = e^x \frac{\cos(e^x)}{\sin(e^x)} = e^x \cot(e^x)

A: $\cot(e^x)$ is incorrect. This happens if the last step of the chain rule (differentiating $e^x$ ) is missed.
B: $e^x \tan(e^x)$ is incorrect. This uses the wrong trigonometric identity, perhaps from misremembering the derivative of $\ln(\sin u)$ .
C: $\frac{e^x}{\sin(e^x)}$ is incorrect. This happens if the middle step of the chain rule (differentiating $\sin(v)$ ) is missed.
D: $e^x \cot(e^x)$ is correct.

Question 14

Find the derivative of $g(x) = \cos^3(2x)$ .

$-3\sin(2x)\cos^2(2x)$
$6\sin(2x)\cos^2(2x)$
$-6\sin(2x)\cos^2(2x)$ (correct answer)
$-6\sin^3(2x)$

Explanation: The function is

g(x) = (\cos(2x))^3

. This requires applying the chain rule twice. Let the outer function be

u^3

, the middle function be

\cos(v)

, and the inner function be

v = 2x

. The derivative is found by differentiating layer by layer from the outside in:

Derivative of the power $u^3$ is $3u^2$ : $3(\cos(2x))^2$
Derivative of the cosine function $\cos(2x)$ is $-\sin(2x)$
Derivative of the inner function $2x$ is $2$

Multiply these together:

g'(x) = 3(\cos(2x))^2 \cdot (-\sin(2x)) \cdot 2

g'(x) = -6\sin(2x)\cos^2(2x)

A: $-3\sin(2x)\cos^2(2x)$ is incorrect. This happens when the derivative of the innermost function $2x$ (which is 2) is forgotten.
B: $6\sin(2x)\cos^2(2x)$ is incorrect. This is a sign error, forgetting the negative sign from the derivative of $\cos(u)$ .
C: $-6\sin(2x)\cos^2(2x)$ is correct.
D: $-6\sin^3(2x)$ is incorrect. This confuses which function is which in the chain rule application.

Question 15

Let $f(x) = \sqrt{\sin(x) + \cos(x)}$ . Find the value of $f'(\frac{\pi}{4})$ .

0 (correct answer)
$\frac{1}{2}$
$\frac{\sqrt{2}}{2}$
$2^{1/4}$

Explanation: First, write the function using a power:

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

. Now, use the chain rule.

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

f'(x) = \frac{\cos x - \sin x}{2\sqrt{\sin x + \cos x}}

Now, substitute

x = \frac{\pi}{4}

. At this value,

\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}

and

\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}

f'(\frac{\pi}{4}) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{2\sqrt{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}}

f'(\frac{\pi}{4}) = \frac{0}{2\sqrt{\sqrt{2}}} = 0

A: 0 is correct.
B: $\frac{1}{2}$ is incorrect. This might be a guess or result from a calculation error.
C: $\frac{\sqrt{2}}{2}$ is incorrect. This is the value of $\cos(\frac{\pi}{4})$ and $\sin(\frac{\pi}{4})$ .
D: $2^{1/4}$ is incorrect. This is the value of the original function $f(\frac{\pi}{4}) = \sqrt{\sqrt{2}} = (2^{1/2})^{1/2} = 2^{1/4}$ , not its derivative.

Question 16

Let $f(x) = x^2 e^{3x}$ . Find $f'(x)$ .

$f'(x) = 6x e^{3x}$
$f'(x) = e^{3x}(2x + x^2)$
$f'(x) = e^{3x}(2x + 3x^2)$ (correct answer)
$f'(x) = e^{3x}(2 + 3x)$

Explanation: To find the derivative of

f(x) = x^2 e^{3x}

, we must use the product rule,

(uv)' = u'v + uv'

, where

u = x^2

and

v = e^{3x}

. First, find the derivatives of

u

and

v

u' = 2x

To find

v'

, we need the chain rule. The derivative of

e^{3x}

e^{3x} \cdot 3 = 3e^{3x}

. Now, apply the product rule:

f'(x) = (2x)(e^{3x}) + (x^2)(3e^{3x})

Finally, factor out the common term

xe^{3x}

for the simplified form, but the answer choices have

e^{3x}

factored out.

f'(x) = e^{3x}(2x + 3x^2)

A: $f'(x) = 6x e^{3x}$ is incorrect. This results from incorrectly multiplying the derivatives of the two parts, i.e., $(2x)(3e^{3x})$ .
B: $f'(x) = e^{3x}(2x + x^2)$ is incorrect. This happens if the chain rule is forgotten when differentiating $e^{3x}$ , treating its derivative as just $e^{3x}$ .
C: $f'(x) = e^{3x}(2x + 3x^2)$ is correct.
D: $f'(x) = e^{3x}(2 + 3x)$ is incorrect. This would arise from incorrectly factoring $x$ out of $2x + 3x^2$ .

Question 17

Find the derivative of $g(x) = \cos^3(2x)$ .

$-3\sin(2x)\cos^2(2x)$
$6\sin(2x)\cos^2(2x)$
$-6\sin(2x)\cos^2(2x)$ (correct answer)
$-6\sin^3(2x)$

Explanation: The function is

g(x) = (\cos(2x))^3

. This requires applying the chain rule twice. Let the outer function be

u^3

, the middle function be

\cos(v)

, and the inner function be

v = 2x

. The derivative is found by differentiating layer by layer from the outside in:

Derivative of the power $u^3$ is $3u^2$ : $3(\cos(2x))^2$
Derivative of the cosine function $\cos(2x)$ is $-\sin(2x)$
Derivative of the inner function $2x$ is $2$

Multiply these together:

g'(x) = 3(\cos(2x))^2 \cdot (-\sin(2x)) \cdot 2

g'(x) = -6\sin(2x)\cos^2(2x)

A: $-3\sin(2x)\cos^2(2x)$ is incorrect. This happens when the derivative of the innermost function $2x$ (which is 2) is forgotten.
B: $6\sin(2x)\cos^2(2x)$ is incorrect. This is a sign error, forgetting the negative sign from the derivative of $\cos(u)$ .
C: $-6\sin(2x)\cos^2(2x)$ is correct.
D: $-6\sin^3(2x)$ is incorrect. This confuses which function is which in the chain rule application.

Question 18

If $f(x) = \ln(\sqrt{x^2+4})$ , what is $f'(x)$ ?

$\frac{1}{2(x^2+4)}$
$\frac{x}{x^2+4}$ (correct answer)
$\frac{x}{\sqrt{x^2+4}}$
$\frac{2x}{\sqrt{x^2+4}}$

Explanation: A helpful first step is to use the properties of logarithms to simplify the function:

f(x) = \ln((x^2+4)^{1/2}) = \frac{1}{2}\ln(x^2+4)

Now, differentiate using the chain rule. The derivative of

\ln(u)

\frac{u'}{u}

. Here,

u = x^2+4

, so

u' = 2x

f'(x) = \frac{1}{2} \cdot \frac{2x}{x^2+4} = \frac{x}{x^2+4}

Alternatively, without simplifying first, we can use the chain rule on

f(x) = \ln(\sqrt{x^2+4})

f'(x) = \frac{1}{\sqrt{x^2+4}} \cdot \frac{d}{dx}(\sqrt{x^2+4})

= \frac{1}{\sqrt{x^2+4}} \cdot \frac{1}{2\sqrt{x^2+4}} \cdot (2x)

= \frac{2x}{2(x^2+4)} = \frac{x}{x^2+4}

A: $\frac{1}{2(x^2+4)}$ is incorrect. This results from forgetting the $2x$ from the chain rule when differentiating $x^2+4$ .
B: $\frac{x}{x^2+4}$ is correct.
C: $\frac{x}{\sqrt{x^2+4}}$ is incorrect. This might result from errors in applying the chain rule for the square root and logarithm functions, possibly missing a factor.
D: $\frac{2x}{\sqrt{x^2+4}}$ is incorrect. This comes from finding the derivative of the inside function, $\sqrt{x^2+4}$ , but failing to apply the derivative of the outer $\ln(u)$ function.

Question 19

The function $f$ is given by $f(x) = \frac{e^x}{e^x + 1}$ . Find $f'(x)$ .

$\frac{-e^x}{(e^x+1)^2}$
1
0
$\frac{e^x}{(e^x+1)^2}$ (correct answer)

Explanation: Use the quotient rule,

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

, with

u = e^x

and

v = e^x + 1

. First, find the derivatives of

u

and

v

u' = e^x

v' = e^x

Apply the quotient rule:

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

Simplify the numerator:

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

f'(x) = \frac{e^x}{(e^x+1)^2}

A: $\frac{-e^x}{(e^x+1)^2}$ is incorrect. This results from reversing the terms in the numerator of the quotient rule to $uv' - u'v$ .
B: 1 is incorrect. This might arise from incorrectly differentiating the numerator and denominator separately and then dividing: $\frac{e^x}{e^x} = 1$ .
C: 0 is incorrect. This might arise from an algebraic error in the numerator, such as $e^x(e^x+1) - e^x(e^x+1) = 0$ .
D: $\frac{e^x}{(e^x+1)^2}$ is correct.

Question 20

Let $h(x) = x \cos(2x^2)$ . Find $h'(x)$ .

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

Explanation: We use the product rule,

(uv)' = u'v + uv'

, with

u = x

and

v = \cos(2x^2)

u' = 1

To find

v'

, we use the chain rule. The derivative of

\cos(w)

-\sin(w) \cdot w'

. Here,

w = 2x^2

, so

w' = 4x

v' = -\sin(2x^2) \cdot 4x = -4x\sin(2x^2)

Now, apply the product rule:

h'(x) = (1)(\cos(2x^2)) + (x)(-4x\sin(2x^2))

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

A: $\cos(2x^2) - 4x^2\sin(2x^2)$ is correct.
B: $\cos(2x^2) + 4x^2\sin(2x^2)$ is incorrect. This is a sign error from the derivative of $\cos(w)$ , treating it as $\sin(w)$ .
C: $\cos(2x^2) - 2x^2\sin(2x^2)$ is incorrect. This results from an error in the chain rule for $v$ , differentiating $2x^2$ as $2x$ instead of $4x$ .
D: $-4x \sin(2x^2)$ is incorrect. This is only the derivative of the second part, $\cos(2x^2)$ , and ignores the product rule.