How do you differentiate f(x)=csc(ln(1-e^x))  using the chain rule?

Dec 28, 2015

Recalling the chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$, we can rename $u = \ln \left(v\right)$ and $v = 1 - {e}^{x}$.

Explanation:

Doing it step-by-step, separately (which is not necessarilt advised, but this is a quite simple function, so it is ok), we'll have that $f \left(x\right) = \csc \left(u\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = - \csc u \cot u$

$\frac{\mathrm{du}}{\mathrm{dv}} = \frac{1}{v}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = - {e}^{x}$

Aggregating and substituting $u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \csc \left(\ln \left(v\right)\right) \cot \left(\ln v\right) \left(\frac{1}{v}\right) \left(- {e}^{x}\right)$

Substituting $v$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} \csc \left(\ln \left(1 - {e}^{x}\right)\right) \cot \left(\ln \left(1 - {e}^{x}\right)\right)}{1 - {e}^{x}}$