# What is the implicit derivative of 1=cosx/y?

Nov 21, 2015

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

#### Explanation:

If we differentiate implicitly and use the quotient rule, we see that:

$\frac{d}{\mathrm{dx}} \left[1 = \cos \frac{x}{y}\right]$

$0 = \frac{\stackrel{\text{-sinx"overbrace(d/dx[cosx])*y-cosx*stackrel"dy/dx}}{\overbrace{\frac{d}{\mathrm{dx}} \left[y\right]}}}{y} ^ 2$

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

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

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