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 $\frac{dy}{dx}$

$\sqrt{3y-1}=5xy$

Short answer

$\frac{dy}{dx}=\frac{10y\sqrt{3y-1}}{3-10x\sqrt{3y-1}}$

Step-by-step solution

Step 1. Given Information: 

Given equation:$\sqrt{3y-1}=5xy$

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

$$\begin{gathered} \frac{d}{dx}\sqrt{3y-1}=\frac{d}{dx}\left(5xy\right) \\ \frac{d}{dx}{(3y-1)}^{\frac{1}{2}}=\frac{d}{dx}\left(5xy\right) \end{gathered}$$

Using product and chain rule we get

$$\begin{gathered} \frac{1}{2}{(3y-1)}^{\frac{1}{2}-1}\frac{d}{dx}(3y-1)=5x\frac{d}{dx}y+5y\frac{d}{dx}x \\ \frac{1}{2\sqrt{3y-1}}\left(3\frac{dy}{dx}\right)=5x\frac{dy}{dx}+5y \\ \frac{3}{2\sqrt{3y-1}}\frac{dy}{dx}-5x\frac{dy}{dx}=5y \\ \left(\frac{3}{2\sqrt{3y-1}}-5x\right)\frac{dy}{dx}=5y \\ \left(\frac{3-10x\sqrt{3y-1}}{2\sqrt{3y-1}}\right)\frac{dy}{dx}=5y \\ \frac{dy}{dx}=\frac{10y\sqrt{3y-1}}{3-10x\sqrt{3y-1}} \end{gathered}$$