Mathematics, 25.07.2021 01:30 ineedhelp2285

Find y' for the following.

Answers

Answer from: Pauline3607

$\displaystyle y' = \frac{5x - 2xy^2}{2y(x^2 - 3y)}$

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

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle 5x^2 - 2x^2y^2 + 4y^3 - 7 = 0$

Step 2: Differentiate

Implicit Differentiation: $\displaystyle \frac{dy}{dx}[5x^2 - 2x^2y^2 + 4y^3 - 7] = \frac{dy}{dx}[0]$ Rewrite [Derivative Property - Addition/Subtraction]: $\displaystyle \frac{dy}{dx}[5x^2] - \frac{dy}{dx}[2x^2y^2] + \frac{dy}{dx}[4y^3] - \frac{dy}{dx}[7] = \frac{dy}{dx}[0]$ Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle 5\frac{dy}{dx}[x^2] - 2\frac{dy}{dx}[x^2y^2] + 4\frac{dy}{dx}[y^3] - \frac{dy}{dx}[7] = \frac{dy}{dx}[0]$ Basic Power Rule [Product Rule, Chain Rule]: $\displaystyle 10x - 2 \Big( \frac{d}{dx}[x^2]y^2 + x^2\frac{d}{dx}[y^2] \Big) + 12y^2y' - 0 = 0$ Basic Power Rule [Chain Rule]: $\displaystyle 10x - 2 \Big( 2xy^2 + x^22yy' \Big) + 12y^2y' - 0 = 0$ Simplify: $\displaystyle 10x - 4xy^2 - 4x^2yy' + 12y^2y' = 0$ Isolate y' terms: $\displaystyle -4x^2yy' + 12y^2y' = 4xy^2 - 10x$ Factor: $\displaystyle y'(-4x^2y + 12y^2) = 4xy^2 - 10x$ Isolate y': $\displaystyle y' = \frac{4xy^2 - 10x}{-4x^2y + 12y^2}$ Simplify: $\displaystyle y' = \frac{5x - 2xy^2}{2y(x^2 - 3y)}$

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

Unit: Differentiation

Book: College Calculus 10e

Answer from: Quest

no this is a nonlinear negative