Question

Find y' for the following.​

Answer 1

Answer:

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

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative 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:

1. f(x) = cxⁿ
2. f’(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

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