Question

If f(x) = 9x10 tan−1x, find f '(x).

90x9tan−1x + 9x10 1 over the quantity 1 plus x squared
90x9tan−1x + 9x10tan−2x
90x9tan−1x − 9x10 1 over the square root of the quantity 1 minus x squared
90x9tan−1x + 9x10 1 over the square root of the quantity 1 minus x squared
those are my choices

Answer 1

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:                                                           $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$  

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

Step 2: Differentiate

1. [Function] Derivative Rule [Product Rule]:                                                   $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
2. Rewrite [Derivative Property - Multiplied Constant]:                                  $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
3. Basic Power Rule:                                                                                         $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
4. Arctrig Derivative:                                                                                         $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

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

Unit: Differentiation