Question

How do you differentiate `f(x)=(cosx-x)/(sin3x-x^2)` using the quotient rule?

Answer 1

`((-sinx-1)(sin3x-x^2)-(3cos3x-2x)(cosx-x))/(sin3x-x^2)^2`

Explanation:

The quotient rule states that

`d/dx f(x)/g(x) = (f'(x)g(x)-g'(x)f(x))/g^2(x)`

Here,

`f(x) = cosx-x`

so

`f'(x) = d/dx[cosx]-d/dx[x] = -sinx-1`

and

`g(x) = sin3x - x^2`

so

`g'(x) = d/dx[sin3x]-d/dx[x^2] = 3cos3x-2x`

We can put these functions into the formula for the quotient rule:

`d/dx f(x)/g(x) = ((-sinx-1)(sin3x-x^2)-(3cos3x-2x)(cosx-x))/(sin3x-x^2)^2`

which you could expand out to give

`= (-sinxsin3x+x^2sinx-sin3x+x^2-3cosxcos3x+3xcos3x+2xcosx-2x^2)/(sin^2(3x)-2x^2sin3x+x^4)`

though this is more complicated and unnecessary.