# How do you implicitly differentiate -3=xsecy?

Jul 21, 2016

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

#### Explanation:

When we implicitly differentiate a function $f \left(y\right)$, we first differentiate $f \left(y\right)$ with respect to $y$ and then multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$. This comes from chain rule as $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \times \frac{\mathrm{dy}}{\mathrm{dx}}$.

Now using product rule for differentiating $3 = x \sec y$, we get

$0 = x \times \sec y + 1 \times \sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}}$ or

$\sec y \tan y \frac{\mathrm{dy}}{\mathrm{dx}} = - x \sec y$ or

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

= $\frac{- x \cancel{\sec y}}{\cancel{\sec y} \tan y}$

= $- x \cot y$