Chapter 6.4, Problem 9QR

To determine

Find $\frac{dy}{dx}$

Answer to Problem 9QR

  $\frac{dy}{dx}=\frac{y+1}{2y-x}$

Explanation of Solution

Given information:

  $xy+x=y^2\mathrm{...}\left(1\right)$

Calculation:

The derivative of equation (1) with respect to x.

  $\begin{array}{l}\frac{d}{dx}\left(xy+x\right)=\frac{d}{dx}\left(y^2\right)\\ \Rightarrow x\frac{dy}{dx}+y\frac{dx}{dx}+\frac{d}{dx}\left(x\right)=2y\frac{dy}{dx}\\ \Rightarrow x\frac{dy}{dx}+y+1=2y\frac{dy}{dx}\\ \Rightarrow 2y\frac{dy}{dx}-x\frac{dy}{dx}=y+1\\ \Rightarrow \frac{dy}{dx}\left(2y-x\right)=y+1\\ \Rightarrow \frac{dy}{dx}=\frac{y+1}{2y-x}\end{array}$

Conclusion:

  $\frac{dy}{dx}=\frac{y+1}{2y-x}$.