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

your answer is d 2,500

step-by-step explanation:

The circle graph represents a family’s monthly budget if the total monthly budget is 5,000 how much

Answer from: Quest

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

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