# How do you differentiate f(x)=csc(lnx)  using the chain rule?

Jun 28, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{1}{x} \csc \left(\ln x\right) \cot \left(\ln x\right)$
As $f \left(x\right) = \csc \left(\ln x\right) = \csc \left(g \left(x\right)\right)$, where $g \left(x\right) = \ln x$
As according to chain rule $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$
$\frac{\mathrm{df}}{\mathrm{dx}} = - \csc \left(\ln x\right) \cot \left(\ln x\right) \times \frac{1}{x} = - \frac{1}{x} \csc \left(\ln x\right) \cot \left(\ln x\right)$