Mathematics, 03.11.2019 21:31 student679

What is the third derivative of tan x?

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Answer from: IamHowToBasic

$\displaystyle y''' = 2 \sec^2 (x) \bigg( 2\tan^2 (x) + \sec^2 (x) \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

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

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

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

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

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

$\displaystyle y = \tan x$

Step 2: Differentiate

Trigonometric Differentiation: $\displaystyle y' = \sec^2 (x)$ Basic Power Rule [Derivative Rule - Chain Rule]: $\displaystyle y'' = 2 \sec (x) \cdot \frac{d}{dx}[\sec (x)]$ Trigonometric Differentiation: $\displaystyle y'' = 2 \sec (x) \cdot \sec (x) \tan (x)$ Simplify: $\displaystyle y'' = 2 \sec^2 (x) \tan (x)$ Derivative Rule [Product Rule]: $\displaystyle y''' = \frac{d}{dx}[2 \sec^2 (x)] \tan (x) + 2 \sec^2 (x) \frac{d}{dx}[\tan (x)]$ Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle y''' = 2 \frac{d}{dx}[\sec^2 (x)] \tan (x) + 2 \sec^2 (x) \frac{d}{dx}[\tan (x)]$ Trigonometric Differentiation: $\displaystyle y''' = 2 \frac{d}{dx}[\sec^2 (x)] \tan (x) + 2 \sec^2 (x) \cdot \sec^2 (x)$ Basic Power Rule [Derivative Rule - Chain Rule]: $\displaystyle y''' = 2 \big( 2 \sec (x) \big) \frac{d}{dx}[\sec (x)] \tan (x) + 2 \sec^2 (x) \cdot \sec^2 (x)$ Trigonometric Differentiation: $\displaystyle y''' = 2 \big( 2 \sec (x) \big) \big( \sec (x) \tan (x) \big) \tan (x) + 2 \sec^2 (x) \cdot \sec^2 (x)$ Simplify: $\displaystyle y''' = 4 \sec^2 (x) \tan^2 (x) + 2 \sec^4 (x)$ Factor: $\displaystyle y''' = 2 \sec^2 (x) \bigg( 2\tan^2 (x) + \sec^2 (x) \bigg)$

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

Unit: Differentiation

Answer from: Quest

The answer is 8 apex

Answer from: Quest

$119.20

step-by-step explanation:

$372.50 - 60% off is $149. $149-20% off is $119.20

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