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

1 Answer
Dec 3, 2015

#f'(x) = - sin(x)#

Explanation:

You can always compute the derivative using the quotient rule, but here a simpler approach is possible.

Let's use the identity

#sin^2 x + cos^2 x = 1 <=> sin^2 x = 1 - cos^2 x#

and the rule

#a^2 - b^2 = (a+b)(a-b)#

to simplify your function:

#f(x) = (sin^2 x)/(1 - cos x) = (1 - cos^2 x )/(1 - cos x ) = ((1 - cos x )(1 + cos x)) / (1 - cos x )#

#color(white)(xx) = (cancel((1 - cos x ))(1 + cos x)) / cancel((1 - cos x )) = 1 + cos x#

Now, it's much easier to compute the derivative!

#f'(x) = - sin(x)#