Question

How do you differentiate `f(x)= ( x sinx )/ ( x - 6)` using the quotient rule?

Answer 1

`f'(x)=((x^2-6x)cosx-6sinx)/(x-6)^2`

Explanation:

The quotient rule states that

`f'(x)=((x-6)color(red)(d/dx[xsinx])-xsinxcolor(blue)(d/dx[x-6]))/(x-6)^2`

Now, find both the internal derivatives.

The first requires product rule:

`d/dx[xsinx]=sinxd/dx[x]+xd/dx[sinx]=color(red)(sinx+xcosx`

The second is very simple.

`d/dx[x-6]=color(blue)1`

Plug these back in.

`f'(x)=((x-6)(sinx+xcosx)-xsinx)/(x-6)^2`

Simplify.

`f'(x)=(xsinx+x^2cosx-6sinx-6xcosx-xsinx)/(x-6)^2`

`f'(x)=((x^2-6x)cosx-6sinx)/(x-6)^2`