# How do you find (dy)/(dx) given y^2tanx=x?

Jan 30, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {y}^{2} {\sec}^{2} x}{2 y \tan x}$

#### Explanation:

We need to use the chain rule when differentiating a function of y since $y$ itself is a function of $x$. First, differentiate that function of $y$ with respect to $y$, then multiply by the derivative of $y$ (with respect to $x$).

Then, we need to use the product rule: if $a , b$ functions of $x$, then $\left(a b\right) ' = a ' b + a b '$.

Assuming $y$ is a function of $x$, differentiate both sides:

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} \tan x + {y}^{2} {\sec}^{2} x = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y \tan x\right) = 1 - {y}^{2} {\sec}^{2} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {y}^{2} {\sec}^{2} x}{2 y \tan x}$