Question

What is the second derivative of `f(x)=cot(3x^2-x)`?

Answer 1

First derivative:

Use the chain rule, which states that `d/dx(cot(u))=-csc^2(u)*(du)/dx`

Thus,

`f'(x)=-csc^2(3x^2-x)*d/dx(3x^2-x)`

`=>-(6x-1)csc^2(3x^2-x)`

Second derivative:

Use product rule in conjunction with the chain rule again.

When doing chain rule with the cosecant function squared, the overriding issue will be the exponent, and then the cosecant.

`f''(x)=-csc^2(3x^2-x)d/dx(6x-1)-(6x-1)d/dx(csc^2(3x^2-x))`

Find each derivative.

`d/dx(6x-1)=6`

`d/dx(csc^2(3x^2-x))=2csc(3x^2-x)d/dx(csc(3x^2-x))`

`color(white)(ssss)` Recall that `d/dx(csc(u))=-csc(u)cot(u)*(du)/dx`.

`=>2csc(3x^2-x) * -csc(3x^2-x)cot(3x^2-x) * (6x-1)`

`=>-2(6x-1)csc^2(3x^2-x)cot(3x^2-x)`

Plug both the derivatives back in to find `f''(x)`.

`f''(x)=-6csc^2(3x^2-x)+2(6x-1)^2csc^2(3x^2-x)cot(3x^2-x)`

This can be further simplified, if you want:

`f''(x)=((72x^2-24x+2)cot(3x^2-x)-6)csc^2(3x^2-x)`