Calculus 1 Quiz: Inverse Trig Derivatives

20 questions · exam conditions

0:00

Inverse Trig DerivativesQuestion 1 of 20

Let $f(x) = x^2 \arcsin(x)$ . What is the value of $f'(\frac{1}{2})$ ?

\frac{\pi + \sqrt{3}}{6}

\frac{\pi}{6} + \frac{\sqrt{3}}{12}

\frac{\pi}{3} + \frac{\sqrt{3}}{6}

\frac{\pi + 2\sqrt{3}}{12}

← Back to quizzes

Calculus 1 Quiz

Calculus 1 Quiz: Inverse Trig Derivatives

Practice Inverse Trig 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 Inverse Trig 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

Let $f(x) = x^2 \arcsin(x)$ . What is the value of $f'(\frac{1}{2})$ ?

$\frac{\pi + \sqrt{3}}{6}$ (correct answer)
$\frac{\pi}{6} + \frac{\sqrt{3}}{12}$
$\frac{\pi}{3} + \frac{\sqrt{3}}{6}$
$\frac{\pi + 2\sqrt{3}}{12}$

Explanation: To find the derivative of

f(x)

, we must use the product rule,

(uv)' = u'v + uv'

, with

u = x^2

and

v = \arcsin(x)

. The derivative is

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

. Now, we evaluate this at

x = \frac{1}{2}

. We have

\arcsin(\frac{1}{2}) = \frac{\pi}{6}

. So,

f'(\frac{1}{2}) = 2(\frac{1}{2})\arcsin(\frac{1}{2}) + (\frac{1}{2})^2 \frac{1}{\sqrt{1-(\frac{1}{2})^2}} = (1)(\frac{\pi}{6}) + (\frac{1}{4})\frac{1}{\sqrt{1-\frac{1}{4}}} = \frac{\pi}{6} + \frac{1}{4} \frac{1}{\sqrt{\frac{3}{4}}} = \frac{\pi}{6} + \frac{1}{4} \frac{1}{\frac{\sqrt{3}}{2}} = \frac{\pi}{6} + \frac{1}{4} \frac{2}{\sqrt{3}} = \frac{\pi}{6} + \frac{1}{2\sqrt{3}}

. Rationalizing the denominator gives

\frac{\pi}{6} + \frac{\sqrt{3}}{6} = \frac{\pi + \sqrt{3}}{6}

Question 2

If $f(x) = \arcsin(\ln x)$ , what is the domain of the derivative, $f'(x)$ ?

$(0, e]$
$[1/e, e]$
$(1/e, e)$ (correct answer)
$(0, \infty)$

Explanation: When finding the domain of a derivative, you need to consider two constraints: where the original function is defined, and where the derivative formula itself is valid. For

f(x) = \arcsin(\ln x)

, let's first find

f'(x)

using the chain rule:

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

Now we need both the original function and its derivative to be defined. The function

f(x)

requires

x > 0

(for the natural log) and

-1 \leq \ln x \leq 1

(for the arcsine). Since

\ln x = 1

when

x = e

and

\ln x = -1

when

x = 1/e

, we need

1/e \leq x \leq e

. The derivative has an additional constraint: the denominator

\sqrt{1-(\ln x)^2}

cannot equal zero. This happens when

(\ln x)^2 = 1

, or when

x = e

x = 1/e

. So the derivative exists only when

1/e < x < e

. Choice A)

(0, e]

ignores the arcsine constraint and includes points where

\ln x < -1

. Choice B)

[1/e, e]

includes the endpoints where the derivative is undefined. Choice D)

(0, \infty)

completely ignores the arcsine domain restriction. Choice C)

(1/e, e)

correctly excludes the endpoints where the derivative formula breaks down. Study tip: For composite function derivatives, always check where both the original function exists AND where the derivative formula doesn't have division by zero or other undefined operations.

Question 3

If $h(t) = \frac{\arctan(t)}{1+t^2}$ , what is $h'(1)$ ?

$-\frac{1}{4}$
$\frac{1-\pi/2}{2}$
$\frac{1-\pi}{4}$
$\frac{2-\pi}{8}$ (correct answer)

Explanation: When you encounter a function that's a quotient involving trigonometric or inverse trigonometric functions, you'll need the quotient rule combined with knowledge of derivative formulas. To find

h'(t)

where

h(t) = \frac{\arctan(t)}{1+t^2}

, apply the quotient rule:

\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

Here,

f(t) = \arctan(t)

and

g(t) = 1+t^2

. You need:

$f'(t) = \frac{1}{1+t^2}$ (derivative of arctangent)
$g'(t) = 2t$

Applying the quotient rule:

h'(t) = \frac{\frac{1}{1+t^2} \cdot (1+t^2) - \arctan(t) \cdot 2t}{(1+t^2)^2} = \frac{1 - 2t\arctan(t)}{(1+t^2)^2}

t = 1

h'(1) = \frac{1 - 2(1)\arctan(1)}{(1+1)^2} = \frac{1 - 2 \cdot \frac{\pi}{4}}{4} = \frac{1 - \frac{\pi}{2}}{4} = \frac{2-\pi}{8}

This confirms answer D is correct. Answer A (

-\frac{1}{4}

) likely comes from forgetting the arctangent term entirely. Answer B (

\frac{1-\pi/2}{2}

) makes the right calculation but uses the wrong denominator—forgetting to square

(1+t^2)

in the quotient rule. Answer C (

\frac{1-\pi}{4}

) incorrectly uses

\arctan(1) = \frac{\pi}{2}

instead of

\frac{\pi}{4}

, a common arctangent value mistake. Remember:

\arctan(1) = \frac{\pi}{4}

, and always square the denominator when applying the quotient rule.

Question 4

If $f(x) = \operatorname{arcsec}(e^{2x})$ , what is $f'(x)$ ?

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

Explanation: The derivative of

\operatorname{arcsec}(u)

\frac{1}{|u|\sqrt{u^2-1}} \frac{du}{dx}

. Here,

u=e^{2x}

, so

\frac{du}{dx} = 2e^{2x}

. Since

e^{2x}

is always positive,

|u| = e^{2x}

. Plugging these into the formula gives:

f'(x) = \frac{1}{e^{2x}\sqrt{(e^{2x})^2-1}} \cdot (2e^{2x}) = \frac{2e^{2x}}{e^{2x}\sqrt{e^{4x}-1}}

. The

e^{2x}

terms cancel, leaving

f'(x) = \frac{2}{\sqrt{e^{4x}-1}}

Question 5

The angle $\theta$ of a searchlight, in radians, is given by $\theta(t) = \operatorname{arccot}(\frac{t}{100})$ , where $t$ is time in seconds. Find the rate of change of the angle in radians per second when $t=100$ .

$-\frac{1}{2}$
$\frac{1}{200}$
$-\frac{1}{200}$ (correct answer)
$-\frac{1}{100}$

Explanation: This problem tests your ability to differentiate inverse trigonometric functions, specifically the arccotangent function. When you see a rate of change question involving inverse trig functions, you need to apply the chain rule along with the derivative formula for that specific function. To find the rate of change, you need to differentiate

\theta(t) = \operatorname{arccot}\left(\frac{t}{100}\right)

with respect to time. The derivative of

\operatorname{arccot}(u)

-\frac{1}{1+u^2} \cdot \frac{du}{dt}

. Here,

u = \frac{t}{100}

, so

\frac{du}{dt} = \frac{1}{100}

. Applying the chain rule:

\frac{d\theta}{dt} = -\frac{1}{1+\left(\frac{t}{100}\right)^2} \cdot \frac{1}{100}

When

t = 100

\frac{d\theta}{dt} = -\frac{1}{1+\left(\frac{100}{100}\right)^2} \cdot \frac{1}{100} = -\frac{1}{1+1} \cdot \frac{1}{100} = -\frac{1}{2} \cdot \frac{1}{100} = -\frac{1}{200}

Answer A (

-\frac{1}{2}

) occurs if you forget to apply the chain rule and miss the

\frac{1}{100}

factor. Answer B (

\frac{1}{200}

) results from forgetting the negative sign in the arccotangent derivative formula. Answer D (

-\frac{1}{100}

) happens if you incorrectly evaluate the denominator as just 1 instead of

1+1=2

. Remember that inverse trig derivatives often involve negative signs and reciprocals. Always double-check the derivative formula and carefully apply the chain rule when the input isn't simply

x

Question 6

Given $\arctan(y) = x$ , find $\frac{d^2y}{dx^2}$ in terms of $y$ .

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

Explanation: This problem tests implicit differentiation and the derivatives of inverse trigonometric functions. When you see an equation involving inverse trig functions that you need to differentiate twice, you'll need to use implicit differentiation systematically. Starting with

\arctan(y) = x

, differentiate both sides with respect to

x

. The derivative of

\arctan(y)

\frac{1}{1+y^2} \cdot \frac{dy}{dx}

(using the chain rule), and the derivative of

x

1

. This gives us:

\frac{1}{1+y^2} \cdot \frac{dy}{dx} = 1

Solving for the first derivative:

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

For the second derivative, differentiate

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

with respect to

x

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

Substituting our expression for

\frac{dy}{dx}

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

Option A)

2y

forgets to account for the chain rule when differentiating

y^2

. Option B)

1+y^2

is actually the first derivative, not the second—this is a common error when students lose track of which derivative they're finding. Option D)

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

appears to result from incorrectly applying the quotient rule or confusing this with a different type of implicit differentiation problem. When working with inverse trig functions and implicit differentiation, always keep track of which derivative you're finding and remember that differentiating

y^2

gives

2y \frac{dy}{dx}

, not just

2y

Question 7

For the curve defined by the equation $y + \arctan(xy) = 1$ , find the derivative $\frac{dy}{dx}$ .

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

Explanation: This problem requires implicit differentiation. We differentiate both sides of the equation with respect to

x

\frac{d}{dx}(y + \arctan(xy)) = \frac{d}{dx}(1)

\frac{dy}{dx} + \frac{d}{dx}(\arctan(xy)) = 0

For the

\arctan(xy)

term, we use the chain rule. Let

u=xy

. The derivative is

\frac{1}{1+u^2} \cdot \frac{du}{dx}

. We need the product rule to find

\frac{du}{dx}

\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}

. So, the equation becomes:

\frac{dy}{dx} + \frac{1}{1+(xy)^2} \left(y + x\frac{dy}{dx}\right) = 0

\frac{dy}{dx} + \frac{y}{1+x^2y^2} + \frac{x}{1+x^2y^2}\frac{dy}{dx} = 0

Now, we solve for

\frac{dy}{dx}

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

Multiply the entire equation by

(1+x^2y^2)

to clear the denominator:

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

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

Distractor B results from an error in the product rule when differentiating $xy$ , specifically treating $x$ as a constant.
Distractor C is a sign error in the final step of solving for $\frac{dy}{dx}$ .
Distractor D represents a significant algebraic error when isolating the $\frac{dy}{dx}$ term.

Question 8

Let $h(x) = \arcsin(x^2)$ . Find the value of $h'(\frac{1}{\sqrt{2}})$ .

$\frac{2\sqrt{6}}{3}$ (correct answer)
$\frac{2}{\sqrt{3}}$
$\frac{4\sqrt{2}}{5}$
$-\frac{2\sqrt{6}}{3}$

Explanation: This is a two-step problem: first find the derivative

h'(x)

, then evaluate it at

x = \frac{1}{\sqrt{2}}

. Step 1: Find

h'(x)

using the chain rule. Let

u = x^2

, so

u' = 2x

. The function is

h(u) = \arcsin(u)

h'(x) = \frac{1}{\sqrt{1-u^2}} \cdot u' = \frac{1}{\sqrt{1-(x^2)^2}} \cdot (2x) = \frac{2x}{\sqrt{1-x^4}}

. Step 2: Evaluate

h'(\frac{1}{\sqrt{2}})

h'(\frac{1}{\sqrt{2}}) = \frac{2(\frac{1}{\sqrt{2}})}{\sqrt{1-(\frac{1}{\sqrt{2}})^4}} = \frac{\sqrt{2}}{\sqrt{1 - \frac{1}{4}}} = \frac{\sqrt{2}}{\sqrt{\frac{3}{4}}} = \frac{\sqrt{2}}{\frac{\sqrt{3}}{2}} = \sqrt{2} \cdot \frac{2}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}}

. Rationalizing the denominator gives

\frac{2\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}

Distractor B is the result of forgetting the chain rule factor of $2x$ , calculating only $\frac{1}{\sqrt{1-x^4}}$ at the given point.
Distractor C is the result of incorrectly using the derivative formula for $\arctan(u)$ , which is $\frac{u'}{1+u^2}$ .
Distractor D is the result of incorrectly using the derivative formula for $\arccos(u)$ , which has a negative sign.

Question 9

Let $y = \arctan(x)$ . What is the value of $\frac{d^2y}{dx^2}$ at $x=1$ ?

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

Explanation: When you encounter a problem asking for the second derivative of an inverse trigonometric function, you need to find the first derivative, then differentiate again using appropriate rules. Starting with

y = \arctan(x)

, let's find the first derivative. The derivative of

\arctan(x)

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

. To find the second derivative, we differentiate

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

using the chain rule. We can rewrite this as

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

, so:

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

x = 1

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

Answer choice A) gives

-1

, which you might get if you forgot to apply the chain rule correctly when differentiating

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

and missed the factor of

2x

. Answer choice B) gives

\frac{1}{2}

, which is the correct magnitude but wrong sign—this happens if you drop the negative sign from the power rule. Answer choice C) gives

1

, which might result from confusing this with the first derivative evaluated at

x = 0

. Remember that second derivatives of inverse trig functions often involve the chain rule twice. Always double-check your sign and don't forget the factor that comes from differentiating the inner function.

Question 10

Evaluate the limit: $\lim_{h\to 0} \frac{\arctan(1+h) - \frac{\pi}{4}}{h}$

$\frac{\pi}{4}$
$\frac{1}{2}$ (correct answer)
$1$
$2$

Explanation: When you see a limit of the form

\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}

, you're looking at the definition of a derivative. This limit represents

f'(a)

, where

f(x) = \arctan(x)

and

a = 1

. Since

\arctan(1) = \frac{\pi}{4}

, our limit becomes

\lim_{h\to 0} \frac{\arctan(1+h) - \arctan(1)}{h}

, which is exactly the derivative of

\arctan(x)

x = 1

. The derivative of

\arctan(x)

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

. At

x = 1

, this gives us

\frac{1}{1+1^2} = \frac{1}{2}

. Therefore, the answer is B)

\frac{1}{2}

. Let's see why the other options are wrong. Choice A)

\frac{\pi}{4}

might tempt you because that's the value of

\arctan(1)

, but remember we're finding the derivative (rate of change), not the function value itself. Choice C)

1

could come from forgetting the derivative formula and thinking the slope of

\arctan(x)

is always 1. Choice D)

2

might result from incorrectly computing

1 + 1^2

as the numerator instead of the denominator in the derivative formula. The key strategy here is recognizing the derivative definition pattern. Whenever you see

\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}

, immediately think "this is

f'(a)

" and find the derivative of the function, then evaluate it at the given point.

Question 11

If $f(x) = \frac{\arccot(x)}{x}$ , what is the value of $f'(1)$ ?

$\frac{\pi-2}{4}$
$-\frac{2+\pi}{4}$ (correct answer)
$\frac{2-\pi}{4}$
$-\frac{1}{2}$

Explanation: When you encounter a function that's a quotient involving inverse trigonometric functions, you'll need the quotient rule combined with derivatives of inverse trig functions. To find

f'(x)

where

f(x) = \frac{\arccot(x)}{x}

, apply the quotient rule:

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

The key derivative you need is

\frac{d}{dx}[\arccot(x)] = -\frac{1}{1+x^2}

. Substituting this:

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

x = 1

f'(1) = \frac{-\frac{1}{2} - \arccot(1)}{1}

Since

\arccot(1) = \frac{\pi}{4}

f'(1) = -\frac{1}{2} - \frac{\pi}{4} = -\frac{2}{4} - \frac{\pi}{4} = -\frac{2+\pi}{4}

This confirms answer B. Answer A gives

\frac{\pi-2}{4}

, which would result from incorrectly adding instead of subtracting terms, or mixing up signs. Answer C gives

\frac{2-\pi}{4}

, suggesting you forgot the overall negative sign from the quotient rule calculation. Answer D gives

-\frac{1}{2}

, which you'd get if you completely ignored the

\arccot(x)

term in the derivative. Study tip: Always memorize that

\frac{d}{dx}[\arccot(x)] = -\frac{1}{1+x^2}

and

\arccot(1) = \frac{\pi}{4}

. When using the quotient rule with inverse trig functions, double-check your signs at each step.

Question 12

Let $f(x) = \arctan(x) + \arctan(\frac{1}{x})$ for $x \neq 0$ . Find $f'(x)$ .

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

Explanation: When you encounter a function involving inverse trigonometric expressions like this, consider whether there might be a special relationship that simplifies the derivative unexpectedly. To find

f'(x)

, you need to differentiate each term separately. For

\arctan(x)

, the derivative is

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

. For

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

, you'll need the chain rule: the derivative of

\arctan(u)

\frac{1}{1+u^2} \cdot u'

, where

u = \frac{1}{x}

and

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

. This gives you

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

. Therefore,

f'(x) = \frac{1}{1+x^2} + (-\frac{1}{1+x^2}) = 0

. Choice A (

\frac{\pi}{2}

) represents a constant that might appear if you incorrectly tried to evaluate the original function rather than differentiate it. Choice B (

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

) would result from incorrectly assuming both terms have the same derivative. Choice C suggests you made sign errors or confused the derivatives of different inverse trig functions. The key insight here is recognizing when terms cancel completely. Always work through chain rule problems systematically, and don't assume complex-looking functions can't have simple derivatives. Sometimes the most elegant mathematical relationships produce surprisingly clean results.

Question 13

If $g(x) = x \arcsin(3x)$ , then $g'(x)$ is:

$\arcsin(3x) + \frac{3x}{\sqrt{1-9x^2}}$ (correct answer)
$\arcsin(3x) + \frac{x}{\sqrt{1-9x^2}}$
$\frac{3}{\sqrt{1-9x^2}}$
$\arcsin(3x) + \frac{3x}{1+9x^2}$

Explanation: This problem requires the product rule and the chain rule. The product rule is

(uv)' = u'v + uv'

. Here, let

u=x

and

v=\arcsin(3x)

. First,

u' = 1

. Second, to find

v'

, we use the chain rule on

\arcsin(3x)

. The derivative of

\arcsin(w)

\frac{1}{\sqrt{1-w^2}}

. Let

w=3x

, so

w'=3

. Thus,

v' = \frac{1}{\sqrt{1-(3x)^2}} \cdot 3 = \frac{3}{\sqrt{1-9x^2}}

. Applying the product rule:

g'(x) = (1) \cdot \arcsin(3x) + x \cdot \left(\frac{3}{\sqrt{1-9x^2}}\right) = \arcsin(3x) + \frac{3x}{\sqrt{1-9x^2}}

Distractor B results from forgetting the chain rule when differentiating $\arcsin(3x)$ , missing the factor of 3.
Distractor C is the result of only differentiating the $\arcsin(3x)$ part and multiplying by $x$ , but completely ignoring the product rule structure.
Distractor D incorrectly uses the derivative formula for $\arctan(w)$ instead of $\arcsin(w)$ .

Question 14

What is an equation of the line tangent to the graph of $f(x) = \arcsec(2x)$ at the point where $x = \frac{1}{\sqrt{3}}$ ?

$y = 3x - \sqrt{3} + \frac{\pi}{3}$
$y = \frac{\sqrt{3}}{2}x + \frac{\pi}{6} - \frac{1}{2}$
$y = 3x - \sqrt{3} + \frac{\pi}{6}$ (correct answer)
$y = 3x + \frac{\pi}{6}$

Explanation: When finding the equation of a tangent line to any function, you need two key pieces: a point on the curve and the slope at that point. For inverse trigonometric functions like arcsecant, this means carefully applying the derivative formula and evaluating at the given x-value. First, find the y-coordinate when

x = \frac{1}{\sqrt{3}}

. Since

f(x) = \arcsec(2x)

, we have

f\left(\frac{1}{\sqrt{3}}\right) = \arcsec\left(\frac{2}{\sqrt{3}}\right) = \arcsec\left(\frac{2\sqrt{3}}{3}\right) = \frac{\pi}{6}

. So our point is

\left(\frac{1}{\sqrt{3}}, \frac{\pi}{6}\right)

. Next, find the slope using the derivative. The derivative of

\arcsec(u)

\frac{1}{|u|\sqrt{u^2-1}}

, so

f'(x) = \frac{2}{|2x|\sqrt{(2x)^2-1}} = \frac{1}{|x|\sqrt{4x^2-1}}

. At

x = \frac{1}{\sqrt{3}}

f'\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{\frac{1}{\sqrt{3}}\sqrt{\frac{4}{3}-1}} = \frac{\sqrt{3}}{\sqrt{\frac{1}{3}}} = 3

. Using point-slope form:

y - \frac{\pi}{6} = 3\left(x - \frac{1}{\sqrt{3}}\right)

, which simplifies to

y = 3x - \sqrt{3} + \frac{\pi}{6}

. This matches choice C. Choice A has

\frac{\pi}{3}

instead of

\frac{\pi}{6}

, likely from incorrectly evaluating the arcsecant. Choice B has the wrong slope and completely different form. Choice D omits the

-\sqrt{3}

term that comes from distributing the slope through the point-slope formula. Remember: with inverse trig derivatives, double-check your algebra carefully—the expressions involve nested radicals and fractions that are easy to miscompute.

Question 15

Let $f(x) = \arccsc(\ln x)$ . What is the value of $f'(e^2)$ ?

$\frac{1}{2e^2\sqrt{3}}$
The derivative does not exist.
$-\frac{1}{2e^2\sqrt{3}}$ (correct answer)
$-\frac{1}{2\sqrt{3}}$

Explanation: When you encounter the composition of inverse trigonometric functions with other functions, you need to apply the chain rule carefully and know the derivatives of inverse trig functions. To find

f'(x)

where

f(x) = \arccsc(\ln x)

, use the chain rule. The derivative of

\arccsc(u)

\frac{-1}{|u|\sqrt{u^2-1}}

, and here

u = \ln x

. So

f'(x) = \frac{-1}{|\ln x|\sqrt{(\ln x)^2-1}} \cdot \frac{d}{dx}[\ln x] = \frac{-1}{|\ln x|\sqrt{(\ln x)^2-1}} \cdot \frac{1}{x}

x = e^2

, we have

\ln(e^2) = 2

. Since

2 > 0

, we have

|\ln(e^2)| = 2

. Therefore:

f'(e^2) = \frac{-1}{2\sqrt{2^2-1}} \cdot \frac{1}{e^2} = \frac{-1}{2\sqrt{3}} \cdot \frac{1}{e^2} = \frac{-1}{2e^2\sqrt{3}}

This confirms answer (C). (A) gives the positive version, which misses the negative sign in the arccosecant derivative formula. (B) is incorrect because the derivative exists everywhere in the domain of

\arccsc(\ln x)

, which requires

|\ln x| \geq 1

. Since

\ln(e^2) = 2 > 1

, the derivative exists. (D) correctly has the negative sign but omits the factor of

\frac{1}{e^2}

that comes from the chain rule. Study tip: Always memorize that inverse trig derivatives often have negative signs (arccsc, arccot, arcsec), and remember to apply the chain rule when the input isn't just

x

Question 16

Let $f(x) = \arctan(e^{2x})$ . What is $f'(x)$ ?

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

Explanation: This problem requires the chain rule along with the derivative of the inverse tangent function. Let

u = e^{2x}

. Then

f(u) = \arctan(u)

. The chain rule states that

\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}

. The derivative of

\arctan(u)

with respect to

u

\frac{1}{1+u^2}

. The derivative of

u = e^{2x}

with respect to

x

\frac{du}{dx} = e^{2x} \cdot 2 = 2e^{2x}

. Multiplying these together gives

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

Distractor B is incorrect because it omits the factor of 2 from the chain rule applied to $e^{2x}$ .
Distractor C incorrectly uses the derivative formula for $\arcsin(u)$ instead of $\arctan(u)$ .
Distractor D incorrectly omits the $e^{2x}$ term from the numerator, which arises from the chain rule.

Question 17

Let $g(x) = \arctan(x)$ . What is the value of $g''(1)$ ?

$-\frac{1}{2}$ (correct answer)
$\frac{1}{2}$
$-1$
$-\frac{1}{4}$

Explanation: First, find the first derivative:

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

. To find the second derivative, we use the chain rule on

g'(x)

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

. Finally, we evaluate

g''(x)

x=1

g''(1) = \frac{-2(1)}{(1+1^2)^2} = \frac{-2}{2^2} = \frac{-2}{4} = -\frac{1}{2}

Question 18

For the curve defined by the equation $y \cdot \arccos(x) = x + y^2$ , what is the value of $\frac{dy}{dx}$ at the point $(0, 1)$ ?

$\frac{4}{\pi - 4}$ (correct answer)
$\frac{2}{\pi - 2}$
$-1$
$\frac{4}{4 - \pi}$

Explanation: We use implicit differentiation. Differentiating both sides with respect to

x

gives:

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

. Rearranging to solve for

\frac{dy}{dx}

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

. Factoring out

\frac{dy}{dx}

yields

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

. Thus,

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

. Substituting the point

(x, y) = (0, 1)

and using

\arccos(0) = \frac{\pi}{2}

, we get

\frac{dy}{dx} = \frac{1 + \frac{1}{\sqrt{1-0^2}}}{\frac{\pi}{2} - 2(1)} = \frac{1+1}{\frac{\pi}{2}-2} = \frac{2}{\frac{\pi-4}{2}} = \frac{4}{\pi-4}

Question 19

If $y = \arctan(\sqrt{x^2 - 1})$ for $x > 1$ , what is $\frac{dy}{dx}$ ?

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

Explanation: This requires the chain rule. Let

u = \sqrt{x^2 - 1}

. Then

y = \arctan(u)

. The chain rule states

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

. We have

\frac{dy}{du} = \frac{1}{1+u^2}

and

\frac{du}{dx} = \frac{1}{2\sqrt{x^2-1}}(2x) = \frac{x}{\sqrt{x^2-1}}

. Substituting

u = \sqrt{x^2-1}

into

\frac{dy}{du}

gives

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

. Therefore,

\frac{dy}{dx} = \frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2-1}} = \frac{1}{x\sqrt{x^2-1}}

Question 20

What is the equation of the line tangent to the graph of $f(x) = \operatorname{arccot}(2x)$ at $x = \frac{1}{2}$ ?

$y - \frac{\pi}{4} = -(x - \frac{1}{2})$ (correct answer)
$y - \frac{\pi}{4} = (x - \frac{1}{2})$
$y - \frac{\pi}{2} = -2(x - \frac{1}{2})$
$y - \frac{\pi}{4} = -\frac{1}{2}(x - \frac{1}{2})$

Explanation: First, find the point of tangency. At

x = \frac{1}{2}

y = f(\frac{1}{2}) = \operatorname{arccot}(2 \cdot \frac{1}{2}) = \operatorname{arccot}(1) = \frac{\pi}{4}

. The point is

(\frac{1}{2}, \frac{\pi}{4})

. Next, find the slope by taking the derivative. The derivative of

\operatorname{arccot}(u)

-\frac{1}{1+u^2} \frac{du}{dx}

. Here,

u=2x

, so

\frac{du}{dx}=2

. Thus,

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

. Evaluate the slope at

x=\frac{1}{2}

m = f'(\frac{1}{2}) = -\frac{2}{1+4(\frac{1}{2})^2} = -\frac{2}{1+4(\frac{1}{4})} = -\frac{2}{1+1} = -1

. Using the point-slope form

y - y_1 = m(x-x_1)

, we get

y - \frac{\pi}{4} = -1(x - \frac{1}{2})