AP Calculus AB: Derivatives of Inverse and Inverse Trig Functions (Drill 1)

Question 1. If \( \dfrac{d}{dx}[\arctan x] = \dfrac{1}{1+x^2} \), what is \( \dfrac{d}{dx}[\arctan(3x)] \)?

• A) \( \dfrac{1}{1+9x^2} \)
• B) \( \dfrac{3}{1+9x^2} \) ✓
• C) \( \dfrac{3}{1+3x^2} \)
• D) \( \dfrac{1}{1+3x^2} \)

Explanation: Choice B is correct. By the chain rule: \( \dfrac{d}{dx}[\arctan(3x)] = \dfrac{1}{1+(3x)^2} \cdot 3 = \dfrac{3}{1+9x^2} \). Choice A is incorrect because the student omitted the chain rule factor of 3 from the inner derivative, writing \( \dfrac{1}{1+9x^2} \) instead of \( \dfrac{3}{1+9x^2} \). Choice C is incorrect because the student included the factor of 3 in the numerator but computed \( (3x)^2 = 3x^2 \) instead of \( 9x^2 \) in the denominator. Choice D is incorrect because the student both omitted the chain rule factor of 3 in the numerator and made the same squaring error, writing \( 3x^2 \) instead of \( 9x^2 \) in the denominator.

Question 2. What is \( \dfrac{d}{dx}[\arcsin x] \)?

• A) \( \dfrac{1}{1+x^2} \)
• B) \( -\dfrac{1}{\sqrt{1-x^2}} \)
• C) \( \dfrac{1}{\sqrt{1-x^2}} \) ✓
• D) \( \dfrac{1}{\sqrt{1+x^2}} \)

Explanation: Choice C is correct. The standard formula is \( \dfrac{d}{dx}[\arcsin x] = \dfrac{1}{\sqrt{1-x^2}} \). Choice A is incorrect because \( \dfrac{1}{1+x^2} \) is the derivative of \( \arctan x \), a common formula mix-up between the two inverse trig derivatives. Choice B is incorrect because \( -\dfrac{1}{\sqrt{1-x^2}} \) is the derivative of \( \arccos x \); the derivatives of arcsin and arccos differ only in sign. Choice D is incorrect because the student used \( \sqrt{1+x^2} \) in the denominator, blending the \( 1+x^2 \) structure of the arctan formula with the radical structure of the arcsin formula.

Question 3. Let \( f \) be a differentiable function with \( f(2) = 5 \) and \( f'(2) = 4 \). If \( g = f^{-1} \), what is \( g'(5) \)?

• A) \( 4 \)
• B) \( \dfrac{1}{4} \) ✓
• C) \( \dfrac{1}{5} \)
• D) \( 2 \)

Explanation: Choice B is correct. The inverse function derivative formula gives \( g'(b) = \dfrac{1}{f'(f^{-1}(b))} \). Since \( f(2) = 5 \), we have \( g(5) = 2 \). Therefore \( g'(5) = \dfrac{1}{f'(2)} = \dfrac{1}{4} \). Choice A is incorrect because the student returned \( f'(2) = 4 \) directly, forgetting to take the reciprocal, the most common error with this formula. Choice C is incorrect because the student confused the formula, computing \( \dfrac{1}{f(2)} = \dfrac{1}{5} \) rather than \( \dfrac{1}{f'(g(5))} \). Choice D is incorrect because the student correctly identified \( g(5) = 2 \) but then reported that intermediate value as the final answer instead of evaluating \( \dfrac{1}{f'(2)} \).

Question 4. The function \( f(x) = x^3 + x \) is one-to-one for all real \( x \). If \( g = f^{-1} \), what is \( g'(2) \)?

• A) \( \dfrac{1}{4} \) ✓
• B) \( \dfrac{1}{3} \)
• C) \( 4 \)
• D) \( \dfrac{1}{13} \)

Explanation: Choice A is correct. To apply \( g'(b) = \dfrac{1}{f'(g(b))} \), first find \( g(2) \) by solving \( a^3 + a = 2 \). Testing \( a = 1 \): \( 1 + 1 = 2 \). ✓ So \( g(2) = 1 \). Then \( f'(x) = 3x^2 + 1 \), giving \( f'(1) = 4 \). Therefore \( g'(2) = \dfrac{1}{4} \). Choice B is incorrect because the student used only the \( 3x^2 \) term of \( f'(x) \), forgetting to differentiate the \( +x \) term, and arrived at \( f'(1) = 3 \). Choice C is incorrect because the student returned \( f'(1) = 4 \) without taking the reciprocal, the same "forgot the reciprocal" error that is the primary misconception for this formula type. Choice D is incorrect because the student evaluated \( f'(2) = 3(4) + 1 = 13 \), plugging in the input value \( b = 2 \) directly instead of first finding \( g(2) = 1 \) and evaluating \( f' \) there.

Question 5. Let \( h(x) = \arcsin(x^2) \). Which of the following correctly gives \( h'(x) \) and identifies a value of \( x \) where \( h'(x) \) does not exist?

• A) \( h'(x) = \dfrac{2x}{\sqrt{1-x^4}} \); \( h'(x) \) does not exist at \( x = 1 \) ✓
• B) \( h'(x) = \dfrac{2x}{\sqrt{1-x^2}} \); \( h'(x) \) does not exist at \( x = 1 \)
• C) \( h'(x) = \dfrac{1}{\sqrt{1-x^4}} \); \( h'(x) \) does not exist at \( x = 0 \)
• D) \( h'(x) = \dfrac{2x}{\sqrt{1-x^4}} \); \( h'(x) \) does not exist at \( x = 0 \)

Explanation: Choice A is correct. By the chain rule: \( h'(x) = \dfrac{1}{\sqrt{1-(x^2)^2}} \cdot 2x = \dfrac{2x}{\sqrt{1-x^4}} \). The derivative is undefined where the denominator equals zero: \( 1 - x^4 = 0 \Rightarrow x = \pm 1 \). So \( h'(x) \) does not exist at \( x = 1 \). Choice B is incorrect because the student failed to square the inner function correctly inside the radical, writing \( 1 - x^2 \) instead of \( 1 - (x^2)^2 = 1 - x^4 \), a chain rule error inside the arcsin formula. Choice C is incorrect because the student omitted the chain rule factor \( 2x \) entirely, applying only the outer derivative without multiplying by the inner derivative; additionally, the claim that \( h'(0) \) does not exist is wrong, at \( x = 0 \) the denominator equals \( \sqrt{1-0} = 1 \), so \( h'(0) = 0 \), which is well-defined. Choice D is incorrect because the formula for \( h'(x) \) is right, but the student misidentified \( x = 0 \) as the point where the derivative fails to exist rather than \( x = 1 \).