# How do you differentiate f(x)=csc(e^(x^2-5x))  using the chain rule?

Jul 14, 2016

You differentiate the outer function, multiply by the derivative of the inner one and go on ...

#### Explanation:

The derivative of $\csc \left(u\right)$ with respect to $u$ is $- \cot \left(u\right) \csc \left(u\right)$. In this problem $u = {e}^{{x}^{2} - 5 x}$, Since we want to differentiate with respect to $x$, rather than with respect to $u$, we must multiply the result by $\frac{\mathrm{du}}{\mathrm{dx}}$ (this, after all, is the chain rule). We have

$\frac{\mathrm{du}}{\mathrm{dx}} = \left(2 x - 5\right) {e}^{{x}^{2} - 5 x}$
(where we have applied the chain rule again)
Putting all this together, we get

$\frac{d}{\mathrm{dx}} \left(\csc \left({e}^{{x}^{2} - 5 x}\right)\right) = - \left(2 x - 5\right) {e}^{{x}^{2} - 5 x} \csc \left({e}^{{x}^{2} - 5 x}\right) \cot \left({e}^{{x}^{2} - 5 x}\right)$