Mathematics, 31.03.2020 06:25, kierafisher05

Y^4+5x=21 second derivative

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answered: Jxsolvez

$\displaystyle y'' = \frac{-75}{16y^7}$

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

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y^4 + 5x = 21$

Step 2: Find 1st Derivative

Differentiate [Basic Power Rule, Chain Rule, Derivative Properties]: $\displaystyle 4y^3y' + 5 = 0$ Isolate y' term: $\displaystyle 4y^3y' = -5$ Isolate y': $\displaystyle y' = \frac{-5}{4y^3}$

Step 3: Find 2nd Derivative

Rewrite: $\displaystyle y' = \frac{-5}{4}y^{-3}$ Differentiation [Basic Power Rule, Chain Rule, Derivative Properties]: $\displaystyle y'' = \frac{15}{4}y^{-4}y'$ Rewrite: $\displaystyle y'' = \frac{15}{4y^4}y'$ Substitute in y': $\displaystyle y'' = \frac{15}{4y^4} \bigg( \frac{-5}{4y^3} \bigg)$ Simplify: $\displaystyle y'' = \frac{-75}{16y^7}$

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

Unit: Differentiation

answered: Guest

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step-by-step explanation:

answered: Guest

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step-by-step explanation:

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