# How do you implicitly differentiate 9=sin^2x-cos^2y?

Dec 21, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\sin x \cos x}{\sin y \cos y}$

#### Explanation:

Recall that when implicitly differentiating, any derivatives of terms with $y$ will spit out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term thanks to the chain rule.

$\frac{d}{\mathrm{dx}} \left[9 = {\sin}^{2} x - {\cos}^{2} y\right]$

$0 = 2 \sin x \cdot \frac{d}{\mathrm{dx}} \left[\sin x\right] - 2 \cos y \cdot \frac{d}{\mathrm{dx}} \left[\cos y\right]$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\sin x \cos x}{\sin y \cos y}$