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 $\frac{dy}{dx}$

role="math" localid="1648660297305" $x^{2}y-y^{2}x=x^2+3$

Short answer

$\frac{dy}{dx}=\frac{2x-2xy+y^2}{(x^2-2xy)}$

Step-by-step solution

Step 1. Given Information: 

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

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

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

Using product and chain rule we get

role="math" localid="1648660863023" $$\begin{gathered} x^2\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(x^2\right)-\left(y^2\frac{d}{dx}\right(x)+x\frac{d}{dx}{y}^2)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(3\right) \\ {x}^2\frac{dy}{dx}+y\left(2x\right)-\left(y^2\right(1)+x(2y\left)\frac{dy}{dx}\right)=\left(2x\right)+0 \\ {x}^2\frac{dy}{dx}+2xy-y^2-2xy\frac{dy}{dx}=2x \\ (x^2-2xy)\frac{dy}{dx}=2x-2xy+y^2 \\ \frac{dy}{dx}=\frac{2x-2xy+y^2}{(x^2-2xy)} \end{gathered}$$