Q. 79 - 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 $\frac{dy}{dx}$

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

Step-by-Step Solution

Answer

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

Step 1. Given Information:

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

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 

$$\begin{gathered} \frac{d}{dx}\left(\frac{1}{y}-\frac{1}{x}\right)=\frac{d}{dx}\left(\frac{x^2}{y+1}\right) \\ \frac{d}{dx}\frac{1}{y}-\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^2{(y+1)}^{-1} \\ U\sin g product and chain rule we get \\ \frac{1}{y^2}\frac{dy}{dx}+\frac{1}{x^2}=x^2\frac{d}{dx}{(y+1)}^{-1}+{(y+1)}^{-1}\frac{d}{dx}{x}^2 \\ \frac{1}{y^2}\frac{dy}{dx}+\frac{1}{x^2}=x^2·(-1{(y+1)}^{-2}\frac{dy}{dx})+{(y+1)}^{-1}·2x \\ \frac{1}{y^2}\frac{dy}{dx}+\frac{1}{x^2}=\frac{-x^2}{{(y+1)}^2}\frac{dy}{dx}+\frac{2x}{(y+1)} \\ \frac{1}{y^2}\frac{dy}{dx}+\frac{x^2}{{(y+1)}^2}\frac{dy}{dx}=\frac{2x}{(y+1)}-\frac{1}{x^2} \\ \left(\frac{1}{y^2}+\frac{x^2}{{(y+1)}^2}\right)\frac{dy}{dx}=\frac{2x}{(y+1)}-\frac{1}{x^2} \\ \left(\frac{{(y+1)}^2+x^2{y}^2}{{y^2(y+1)}^2}\right)\frac{dy}{dx}=\frac{2x^3-(y+1)}{x^2(y+1)} \\ \frac{dy}{dx}=\left(\frac{2x^3-(y+1)}{x^2(y+1)}\right)\left(\frac{{y^2(y+1)}^2}{{(y+1)}^2+x^2{y}^2}\right) \\ \frac{dy}{dx}=\frac{y^2(y+1)(2x^3-(y+1\left)\right)}{x^2({(y+1)}^2+x^2{y}^2)} \end{gathered}$$