Q. 77 - 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 $\frac{dy}{dx}$

$\frac{y^2+1}{3y-1}=x$

Step-by-Step Solution

Answer

$\frac{dy}{dx}=\frac{-1}{3}·\frac{{(3y-1)}^2}{7y^2-2y+1}$

Step 1. Given Information:

Given equation: $\frac{y^2+1}{3y-1}=x$

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 

$\frac{d}{dx}(y^2+1){(3y-1)}^{-1}=\frac{d}{dx}x$

Using product and chain rule we get 

$$\begin{gathered} (y^2+1)\frac{d}{dx}{(3y-1)}^{-1}+{(3y-1)}^{-1}\frac{d}{dx}(y^2+1)=\frac{d}{dx}x \\ (y^2+1)\left({-1(3y-1)}^{-2}\right)\frac{d}{dx}(3y-1)+{(3y-1)}^{-1}(2y\frac{d}{dx}y+0)=\frac{d}{dx}x \\ (y^2+1)\left({-1(3y-1)}^{-2}\right)(3\frac{dy}{dx}-0)+{(3y-1)}^{-1}\left(2y\frac{dy}{dx}\right)=1 \\ (-3)\frac{y^2+1}{{(3y-1)}^2}\frac{dy}{dx}+\frac{2y}{3y-1}\frac{dy}{dx}=1 \\ \left((-3)\frac{y^2+1}{{(3y-1)}^2}+\frac{2y}{3y-1}\right)\frac{dy}{dx}=1 \\ \left((-3)\frac{y^2+1+2y(3y-1)}{{(3y-1)}^2}\right)\frac{dy}{dx}=1 \\ \left((-3)\frac{y^2+1+6y^2-2y}{{(3y-1)}^2}\right)\frac{dy}{dx}=1 \\ \left((-3)\frac{7y^2-2y+1}{{(3y-1)}^2}\right)\frac{dy}{dx}=1 \\ \frac{dy}{dx}=\frac{-1}{3}·\frac{{(3y-1)}^2}{7y^2-2y+1} \end{gathered}$$