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

`\displaystyle f'(x) = 90x^9 \tan^(-1)(x) + (9x^(10))/(x^2 + 1)`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (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 (d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)`

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

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

Unit: Differentiation