How do you differentiate f(x)=csc(1/x^2-x)  using the chain rule?

Mar 19, 2016

$f ' \left(x\right) = \left(\frac{2}{x} ^ 3 + 1\right) \csc \left(\frac{1}{x} ^ 2 - x\right) \cot \left(\frac{1}{x} ^ 2 - x\right)$

Explanation:

The derivative of $\csc \left(x\right)$ is:

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

The chain rule states that when differentiating a function inside another function, differentiate the outside function while leaving the inside function intact, and then multiply that by the derivative of the inside function.

When applied to $\csc \left(u\right)$, this can be shown as:

$\frac{d}{\mathrm{dx}} \csc \left(u\right) = - \csc \left(u\right) \cot \left(u\right) \cdot u '$

Here, $u = \frac{1}{x} ^ 2 - x$, so

$u = {x}^{-} 2 - x \text{ "=>" } u ' = - 2 {x}^{-} 3 - 1$

$u ' = - \frac{2}{x} ^ 3 - 1$

Thus,

$\frac{d}{\mathrm{dx}} \csc \left(\frac{1}{x} ^ 2 - x\right) = - \csc \left(\frac{1}{x} ^ 2 - x\right) \cot \left(\frac{1}{x} ^ 2 - x\right) \cdot \left(- \frac{2}{x} ^ 3 - 1\right)$

Distributing the negative,

$= \left(\frac{2}{x} ^ 3 + 1\right) \csc \left(\frac{1}{x} ^ 2 - x\right) \cot \left(\frac{1}{x} ^ 2 - x\right)$