\dfrac{d}{dx} e^{x^2} =

Answer: 2x \cdot e^{x^2} Chain rule: e^{x^2} \cdot 2x.

\dfrac{d}{dx}\ln(x^2 + 1) =

Answer: \dfrac{2x}{x^2 + 1} \dfrac{1}{x^2+1} \cdot 2x = \dfrac{2x}{x^2+1}.

Answer: 20(2x - 1)^9 10(2x-1)^9 \cdot 2 = 20(2x-1)^9.

AP Calculus AB · Topic 3.1

Part of Differentiation: Composite, Implicit, Inverse Functions.(FUN-3.A)

Practice questions

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Sample 1difficulty 2/5
$\dfrac{d}{dx} e^{x^2} =$
- A
  $x^2 e^{x^2}$
- B
  $2x \cdot e^{2x}$
- C
  $2x \cdot e^{x^2}$
  check_circle
- D
  $e^{2x}$
Why
$e^{x^2} \cdot 2x$
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Sample 2difficulty 2/5
$\dfrac{d}{dx}\ln(x^2 + 1) =$
- A
  $\ln(2x)$
- B
  $\dfrac{2x}{\ln(x^2+1)}$
- C
  $\dfrac{2x}{x^2 + 1}$
  check_circle
- D
  $\dfrac{1}{x^2 + 1}$
Why
$\dfrac{1}{x^2+1} \cdot 2x = \dfrac{2x}{x^2+1}$
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Sample 3difficulty 2/5
$\dfrac{d}{dx}(2x - 1)^{10} =$
- A
  $5(2x - 1)^9$
- B
  $20(2x - 1)^9$
  check_circle
- C
  $10(2x - 1)^9$
- D
  $2(2x-1)^{10}$
Why
$10(2x-1)^9 \cdot 2 = 20(2x-1)^9$
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Sample 4difficulty 2/5
$\dfrac{d}{dx}\cos(3x + 1) =$
- A
  $3\cos(3x + 1)$
- B
  $-3\sin(3x + 1)$
  check_circle
- C
  $-\sin(3x + 1)$
- D
  $3\sin(3x + 1)$
Why
$-\sin(3x+1) \cdot 3 = -3\sin(3x+1)$
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Sample 5difficulty 2/5
$\dfrac{d}{dx}\sin(x^3) =$
- A
  $3x^2 \cos(x^3)$
  check_circle
- B
  $3x^2 \sin(x^3)$
- C
  $\cos(3x^2)$
- D
  $\cos(x^3)$
Why
$\cos(x^3) \cdot 3x^2$
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