What is the derivative of $f (x) = \frac{{sin}^{2} x}{1 - cos x}$ $f(x)=(sin^2x)/(1-cosx)$ ?

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Answer 1 · 2015-12-03T10:57:35

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

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!

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

What is the derivative of f(x)=(sin^2x)/(1-cosx)f(x)=sin2x1−cosx f(x)=(sin^2x)/(1-cosx)?