Question

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

Answer 1

`f'(x)=(2/x^3+1)csc(1/x^2-x)cot(1/x^2-x)`

Explanation:

The derivative of `csc(x)` is:

`d/dxcsc(x)=-csc(x)cot(x)`

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(u)`, this can be shown as:

`d/dxcsc(u)=-csc(u)cot(u)*u'`

Here, `u=1/x^2-x`, so

`u=x^-2-x" "=>" "u'=-2x^-3-1`

`u'=-2/x^3-1`

Thus,

`d/dxcsc(1/x^2-x)=-csc(1/x^2-x)cot(1/x^2-x)*(-2/x^3-1)`

Distributing the negative,

`=(2/x^3+1)csc(1/x^2-x)cot(1/x^2-x)`