# What is the derivative of y = ln(cscx)?

Aug 22, 2016

We will use the chain rule to differentiate. Let $y = \ln u$ and $u = \csc x$

The derivative of $\csc x$, by the quotient rule, is $\left(\csc x\right) ' = \frac{- \cos x}{\sin} ^ 2 x = - {\csc}^{2} x \cos x = - \cot x \csc x$

The derivative of $\ln u$ is $\frac{1}{u}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times - \cot x \csc x = \frac{1}{\csc} x \times - \cot x \csc x = \sin x \times \left(- \cot x \times \frac{1}{\sin} x\right) = - \cot x$

Hopefully this helps!

Apr 5, 2018

Alternatively, using logarithm laws, we can say:

$y = \ln \left(\frac{1}{\sin} x\right) = \ln \left(1\right) - \ln \left(\sin x\right) = 0 - \ln \left(\sin x\right) = - \ln \left(\sin x\right)$

It follows by the chain rule that

$y ' = \cos x \cdot - \frac{1}{\sin} x = - \cos \frac{x}{\sin} x = - \cot x$

Hopefully this helps!