# How do you differentiate f(x)=1/sin(e^arctanx) using the chain rule?

Nov 10, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\csc \left({e}^{\arctan \left(x\right)}\right) \cot \left({e}^{\arctan \left(x\right)}\right) {e}^{\arctan \left(x\right)}}{{x}^{2} + 1}$

#### Explanation:

So we have

$y = \frac{1}{\sin} \left({e}^{\arctan \left(x\right)}\right) = \csc \left({e}^{\arctan \left(x\right)}\right)$

While the cosecant has a tabled, standard derivative we can work with the sine too, if you don't remember / don't have a standard derivative table handy.

Let's say that $u = \sin \left({e}^{\arctan \left(x\right)}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} {u}^{- 1} \frac{\mathrm{du}}{\mathrm{dx}}$
dy/dx = -u^(-3/2)d/dx(sin(e^(arctan(x)))

Let's say that $v = {e}^{\arctan \left(x\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {u}^{- \frac{3}{2}} \frac{d}{\mathrm{dv}} \sin \left(v\right) \frac{\mathrm{dv}}{\mathrm{dx}} = - {u}^{- \frac{3}{2}} \cos \left(v\right) \frac{d}{\mathrm{dx}} \left({e}^{\arctan \left(x\right)}\right)$

Let's say that $w = \arctan \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {u}^{- \frac{3}{2}} \cos \left(v\right) \frac{d}{\mathrm{dw}} {e}^{w} \frac{\mathrm{dw}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {u}^{- \frac{3}{2}} \cos \left(v\right) {e}^{w} \frac{d}{\mathrm{dx}} \left(\arctan \left(x\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{u}^{- \frac{3}{2}} \cos \left(v\right) {e}^{w}}{{x}^{2} + 1}$

Putting it all in terms of $x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\csc \left({e}^{\arctan \left(x\right)}\right) \cot \left({e}^{\arctan \left(x\right)}\right) {e}^{\arctan \left(x\right)}}{{x}^{2} + 1}$