Q. 78 - Calculus

Question

Each of the equations in Exercises 69–80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{dy}{dx}$

$3y=5x^2+\sqrt[3]{y-2}$

Step-by-Step Solution

Answer

$\frac{dy}{dx}=\frac{30x{(y-2)}^{{\displaystyle \frac{2}{3}}}}{9{(y-2)}^{\frac{2}{3}}-1}$

Step 1. Given Information:

Given equation: $3y=5x^2+\sqrt[3]{y-2}$

We want to find $\frac{dy}{dx}$ defines y as an implicit function of x by use implicit differentiation.

Step 2. Solution:

Differentiate both sides w.r.t. x  

$$\begin{gathered} \frac{d}{dx}3y=\frac{d}{dx}(5x^2+\sqrt[3]{y-2}) \\ \frac{d}{dx}3y=\frac{d}{dx}5x^2+\frac{d}{dx}\sqrt[3]{y-2} \end{gathered}$$

Using product and chain rule we get 

$$\begin{gathered} 3\frac{dy}{dx}=10x+\frac{1}{3}{(y-2)}^{\frac{1}{3}-1}\frac{d}{dx}(y-2) \\ 3\frac{dy}{dx}=10x+\frac{1}{3{(y-2)}^{{\displaystyle \frac{2}{3}}}}\frac{dy}{dx} \\ 3\frac{dy}{dx}-\frac{1}{3{(y-2)}^{{\displaystyle \frac{2}{3}}}}\frac{dy}{dx}=10x \\ \left(3-\frac{1}{3{(y-2)}^{{\displaystyle \frac{2}{3}}}}\right)\frac{dy}{dx}=10x \\ \left(\frac{9{(y-2)}^{{\displaystyle \frac{2}{3}}}-1}{3{(y-2)}^{{\displaystyle \frac{2}{3}}}}\right)\frac{dy}{dx}=10x \\ \frac{dy}{dx}=\frac{30x{(y-2)}^{{\displaystyle \frac{2}{3}}}}{9{(y-2)}^{\frac{2}{3}}-1} \end{gathered}$$