Q. 80 - 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 $\frac{dy}{dx}$

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

Step-by-Step Solution

Answer

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

Step 1. Given Information:

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

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  

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

Using product and chain rule we get 

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