Question

What is the derivative of `f(x)=(sinx^2)/(1-cosx)`?

Answer 1

Assuming no error in the question, `f'(x)=(2xcos(x^2)-2cos(x)cos(x^2)-sin(x)sin(x^2))/(1-cosx)^2`

Explanation:

Assuming no error in the question.
Apply the quotient rule with numerator `sin(x^2)` and denominator `1-cosx`

`d/dx(sin(x^2)) = cos(x^2) d/dx(x^2)` `" "` `" "` (use the chain rule)

`= 2xcos(x^2)`

And `d/dx(1-cosx) = sinx`

`f(x) = sin(x^2)/(1-cosx)`

We get,

`f'(x) = ([2xcos(x^2)] (1-cosx) - sin(x^2) [sinx])/(1-cosx)^2`

This can be rewritten:

`f'(x)=(2xcos(x^2)-2cos(x)cos(x^2)-sin(x)sin(x^2))/(1-cosx)^2`

If there is an error in the question

If the question is intended to ask for the derivative of

`f(x) = (sinx)^2/(1-cosx) = sin^2x/(1-cosx)`,

then instead of using the quotient rule immediately, it is simpler to apply some trigonometry first.

`sin^2x/(1-cosx) = sin^2x/(1-cosx) * (1+cosx)/(1+cosx)`

`= (sin^2x(1+cosx))/(1-cos^2x)`

`= (sin^2x(1+cosx))/sin^2x`

`= 1+cosx`.

That is, `f(x) = 1+cosx`

So the derivative is `f'(x) = -sinx`.

I think that this is simpler than applying the quotient rule to get,

`f'(x) = (2sinxcosx(1-cosx)-sin^2x(sinx))/(1-cosx)^2`,

and then simplifying to

`f'(x) = (sinx(2cosx-2cos^2x-sin^2x))/(1-cosx)^2`

`= (sinx(2cosx-cos^2x-cos^2x-sin^2x))/(1-cosx)^2`

`= (sinx(2cosx-cos^2x-1))/(1-cosx)^2`

`= (sinx(-1+2cosx-cos^2x))/(1-cosx)^2`

`= -sinx`