# How do you differentiate f(x)=ln(sinx)/cosx using the quotient rule?

Feb 8, 2016

$f ' \left(x\right) = \frac{\cos x \cot x + \sin x \ln \left(\sin x\right)}{\cos} ^ 2 x$

#### Explanation:

The quotient rule states that

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

When this is applied to $f \left(x\right)$, the function at hand, we see that

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

Find each derivative separately:

To differentiate the natural logarithm function, use the chain rule, which for a natural logarithm function states that

$\frac{d}{\mathrm{dx}} \left[\ln \left(k \left(x\right)\right)\right] = \frac{1}{k \left(x\right)} \cdot k ' \left(x\right)$

Here, $k \left(x\right) = \sin x$, so

$\frac{d}{\mathrm{dx}} \left[\ln \left(\sin x\right)\right] = \frac{1}{\sin} x \cdot \frac{d}{\mathrm{dx}} \left[\sin x\right] = \frac{1}{\sin} x \cdot \cos x = \cot x$

As for the other derivative,

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

Plugging these both back in, we see that

$f ' \left(x\right) = \frac{\cos x \left(\cot x\right) - \ln \left(\sin x\right) \left(- \sin x\right)}{\cos} ^ 2 x$

This can be written as

$f ' \left(x\right) = \frac{\cos x \cot x + \sin x \ln \left(\sin x\right)}{\cos} ^ 2 x$

As with many trigonometric functions, this can be rewritten in many ways, including

$f ' \left(x\right) = \sec x \left(\cot x + \tan x \ln \left(\sin x\right)\right)$