# How do you find dy/dx by implicit differentiation given lny=sinx?

Feb 1, 2017

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

#### Explanation:

Differentiate both sides with respect to $x$, keeping in mind that:

$\frac{d}{\mathrm{dx}} f \left(y \left(x\right)\right) = f ' \left(y \left(x\right)\right) \cdot y ' \left(x\right)$

so:

$\frac{d}{\mathrm{dx}} \ln y = \frac{d}{\mathrm{dx}} \sin x$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x$

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

We can note that the original equation allows us also to express $y \left(x\right)$ explicitly:

$\ln y = \sin x \iff y = {e}^{\sin} x$

so that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x {e}^{\sin} x$