Section 12.5 Exercise 24

Find \(\frac{dy}{dx}\) using Implicit Differentiation and Theorem 12.5.3

\[ (3x^2+2y^3)^4=2 \]

Using Implicit Differentiation:

\[ \frac{d}{dx}((3x^2+2y^3)^4) = \frac{d}{dx}(2) \]

\[ 4(3x^2+2y^3)^3(6x+6y^2\frac{dy}{dx}) = 0 \]

\[ 6x+6y^2\frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{x}{y^2} \]

Using Theorem 12.5.3

\[ \frac{dy}{dx} = -\frac{f_x}{f_y} \]

\[ f_x = 4(3x^2+2y^3)^3(6x), \ \ \ \ f_y = 4(3x^2+2y^3)^3(6y^2) \]

\[ \frac{dy}{dx} = -\frac{f_x}{f_y} = - \frac{4(3x^2+2y^3)^3(6x)}{4(3x^2+2y^3)^3(6y^2)} = -\frac{x}{y^2} \]