Practice Problems – Derivatives – AP Calculus

Below are 20 practice problems covering all topics from Module 02A: Derivatives – Concepts & Rules. Topics include: limit definition, power rule, product/quotient rules, chain rule, trig derivatives, implicit differentiation, and inverse trig derivatives. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Power Rule

Find $\displaystyle \frac{d}{dx}(3x^5 - 2x^3 + 4x - 7)$.

Solution:
$\frac{d}{dx}3x^5 = 15x^4$
$\frac{d}{dx}(-2x^3) = -6x^2$
$\frac{d}{dx}(4x) = 4$
$\frac{d}{dx}(-7) = 0$
$\displaystyle \frac{d}{dx}(3x^5 - 2x^3 + 4x - 7) = 15x^4 - 6x^2 + 4$

Problem 02 Product Rule

Find $\displaystyle \frac{d}{dx}(x^2 \sin x)$.

Solution:
$f(x) = x^2$, $g(x) = \sin x$
$f'(x) = 2x$, $g'(x) = \cos x$
$\frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) = 2x\sin x + x^2\cos x$

Problem 03 Quotient Rule

Find $\displaystyle \frac{d}{dx}\left(\frac{x^2}{x+1}\right)$.

Solution:
$f(x) = x^2$, $g(x) = x+1$
$f'(x) = 2x$, $g'(x) = 1$
$\frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$

Problem 04 Chain Rule

Find $\displaystyle \frac{d}{dx}(3x+1)^5$.

Solution:
Outer: $u^5$, Inner: $3x+1$
$\frac{d}{dx}(3x+1)^5 = 5(3x+1)^4 \cdot 3 = 15(3x+1)^4$

Problem 05 Trig Derivative

Find $\displaystyle \frac{d}{dx}\cos(5x)$.

Solution:
$\frac{d}{dx}\cos(5x) = -\sin(5x) \cdot 5 = -5\sin(5x)$

Problem 06 Chain Rule with Trig

Find $\displaystyle \frac{d}{dx}\sin^2(2x)$.

Solution:
$\sin^2(2x) = [\sin(2x)]^2$
$\frac{d}{dx}[\sin(2x)]^2 = 2\sin(2x) \cdot \cos(2x) \cdot 2 = 4\sin(2x)\cos(2x) = 2\sin(4x)$

Problem 07 Exponential Derivative

Find $\displaystyle \frac{d}{dx}e^{x^2}$.

Solution:
$\frac{d}{dx}e^{x^2} = e^{x^2} \cdot 2x = 2x e^{x^2}$

Problem 08 Logarithm Derivative

Find $\displaystyle \frac{d}{dx}\ln(x^2+1)$.

Solution:
$\frac{d}{dx}\ln(x^2+1) = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$

Problem 09 Implicit Differentiation

Find $\displaystyle \frac{dy}{dx}$ for $x^2 + y^2 = 16$.

Solution:
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0$
$2x + 2y \frac{dy}{dx} = 0$
$2y \frac{dy}{dx} = -2x$
$\frac{dy}{dx} = -\frac{x}{y}$

Problem 10 Implicit with Product Rule

Find $\displaystyle \frac{dy}{dx}$ for $x^2 + xy + y^2 = 5$.

Solution:
$2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0$
$(x + 2y)\frac{dy}{dx} = -2x - y$
$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

Problem 11 Inverse Trig

Find $\displaystyle \frac{d}{dx}\arcsin(3x)$.

Solution:
$\frac{d}{dx}\arcsin(3x) = \frac{1}{\sqrt{1-(3x)^2}} \cdot 3 = \frac{3}{\sqrt{1-9x^2}}$

Problem 12 Inverse Trig

Find $\displaystyle \frac{d}{dx}\arctan(x^2)$.

Solution:
$\frac{d}{dx}\arctan(x^2) = \frac{1}{1+(x^2)^2} \cdot 2x = \frac{2x}{1+x^4}$

Problem 13 Chain Rule

Find $\displaystyle \frac{d}{dx}\sqrt{x^2+1}$.

Solution:
$\sqrt{x^2+1} = (x^2+1)^{1/2}$
$\frac{d}{dx}(x^2+1)^{1/2} = \frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+1}}$

Problem 14 Tangent Line

Find the slope of the tangent line to $f(x) = x^2$ at $x = 2$.

Solution:
$f'(x) = 2x$, $f'(2) = 4$
The slope of the tangent line at $x = 2$ is $4$.

Problem 15 Tangent Line Equation

Find the equation of the tangent line to $f(x) = x^2$ at $x = 2$.

Solution:
$f(2) = 4$, $f'(2) = 4$
$y - 4 = 4(x - 2)$ ⇒ $y = 4x - 4$

Problem 16 Inverse Function

If $f(x) = x^3 + 1$, find $(f^{-1})'(2)$.

Solution:
$f^{-1}(2)$: $x^3 + 1 = 2$ ⇒ $x^3 = 1$ ⇒ $x = 1$
$f'(x) = 3x^2$, $f'(1) = 3$
$(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{3}$

Problem 17 Higher Order

Find $f''(x)$ for $f(x) = x^3 - 2x^2 + 5x - 1$.

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

Problem 18 Implicit with Trig

Find $\displaystyle \frac{dy}{dx}$ for $\sin x + \cos y = 1$.

Solution:
$\cos x - \sin y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = \frac{\cos x}{\sin y}$

Problem 19 Chain Rule with Exponentials

Find $\displaystyle \frac{d}{dx}e^{\sin x}$.

Solution:
$\frac{d}{dx}e^{\sin x} = e^{\sin x} \cdot \cos x = e^{\sin x} \cos x$

Problem 20 Chain Rule with Logarithms

Find $\displaystyle \frac{d}{dx}\ln(\ln x)$.

Solution:
$\frac{d}{dx}\ln(\ln x) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}$ (for $x > 1$)

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