Question

Let f be a differentiable function. If h (x) = (2+ f (sin x))", which of the following gives a correct process for finding h' (x) ?

Answer 1

Answer:

$\displaystyle h'(x) = \cos (x) f'(\sin x)[2 + f(\sin x)]'''$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                 $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle h(x) = [2 + f(\sin x)]''$

Step 2: Differentiate

1. Derivative Rule [Chain Rule]:                                                                       $\displaystyle h'(x) = [2 + f(\sin x)]''' [2 + f(\sin x)]'$
2. Rewrite [Derivative Property - Addition/Subtraction]:                                 $\displaystyle h'(x) = [2 + f(\sin x)]''' \bigg[(2)' + [f(\sin x)]' \bigg]$
3. Derivative Rule [Basic Power Rule]:                                                             $\displaystyle h'(x) = [2 + f(\sin x)]''' [f(\sin x)]'$
4. Derivative Rule [Chain Rule]:                                                                       $\displaystyle h'(x) = [2 + f(\sin x)]''' f'(\sin x)(\sin x)'$
5. Trigonometric Differentiation:                                                                       $\displaystyle h'(x) = \cos (x) f'(\sin x)[2 + f(\sin x)]'''$

From here, if you would like, you can take the 3rd derivative of the last piece.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation