# What is the implicit derivative of 1=tanx-y^2?

##### 1 Answer
Nov 21, 2015

$2 \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} ^ 2 \left(x\right)$

#### Explanation:

y is an (unknown) function of x

Derive the left hand side
$\frac{d}{\mathrm{dx}} \left(1\right) = 0$

Derive the right hand side

d/dx(tan(x) - y^2) =d/dx(tan(x) - d/dx(y^2)
Sum rule in differention

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
$\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} \left(x\right)\right)$
$\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right) = \frac{\sin \left(x\right) \cdot \sin \left(x\right) - \left(- \cos \left(x\right)\right) \cdot \cos \left(x\right)}{\cos} {\left(x\right)}^{2}$
$\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right) = \frac{{\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right)$
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
$\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right) = \frac{1}{\cos} ^ 2 \left(x\right)$

$\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 2 \cdot y \cdot \frac{d}{\mathrm{dx}} \left(y\right) = 2 \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
The chain rule.

d/dx(tan(x)) = 1/cos^2(x) - 2ydy/dx #

Left hand side = right hand side

$0 = \frac{1}{\cos} ^ 2 \left(x\right) - 2 \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Rewrite

$2 \cdot y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} ^ 2 \left(x\right)$