# How do you use the quotient rule to differentiate y=cos(x)/ln(x)?

Aug 4, 2014

The quotient rule states:

$\frac{d}{\mathrm{dx}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$

Let $f \left(x\right) = \cos x$, and let $g \left(x\right) = \ln x$.

We know that the derivative of $\cos x$ is $- \sin x$, and that the derivative of $\ln x$ is $\frac{1}{x}$. Therefore, $f ' \left(x\right) = - \sin x$, and $g ' \left(x\right) = \frac{1}{x}$.

Now we may simply plug into the quotient rule formula:

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

And there is our answer. If we would like, we can split this fraction up to make it a bit prettier:

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

This simplifies to:

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