# What is the derivative of ((sinx)^2)/(1-cosx)?

Jul 31, 2015

${y}^{'} = - \sin x$

#### Explanation:

Here's an excellent example of how one simple trigonometric identity can spare you a significant amount of work on this derivative.

More specifically, you can use the fact that

$\textcolor{b l u e}{{\sin}^{x} + {\cos}^{2} x = 1 \implies {\sin}^{2} x = 1 - {\cos}^{2} x}$

$y = \frac{1 - {\cos}^{2} x}{1 - \cos x} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 - \cos x\right)}}} \cdot \left(1 + \cos x\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 - \cos x\right)}}}} = 1 + \cos x$

The derivative of $y$ will thus be

${y}^{'} = \frac{d}{\mathrm{dx}} \left(1 + \cos x\right) = \textcolor{g r e e n}{- \sin x}$

Now, let's assume that you didn't notice you could use this identity to simplify your function.

Since your function can be written as

$y = f \frac{x}{g} \left(x\right)$, with $g \left(x\right) \ne 0$

you can use the quotient rule to differentiate it by

color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2

Using this approach, the derivative of $y$ would be

${y}^{'} = \frac{\left[\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right)\right] \cdot \left(1 - \cos x\right) - {\sin}^{2} x \cdot \frac{d}{\mathrm{dx}} \left(1 - \cos x\right)}{1 - \cos x} ^ 2$

You can find $\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right)$ by using the chain rule to write

${\sin}^{2} x = {u}^{2}$, with $u = \sin x$

This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{2}\right) = \frac{d}{\mathrm{du}} {u}^{2} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({u}^{2}\right) = 2 u \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} x\right) = 2 \sin x \cdot \cos x$

Take this back to your target derivative to get

${f}^{'} = \frac{2 \cdot \sin x \cdot \cos x \cdot \left(1 - \cos x\right) - {\sin}^{2} x \cdot \sin x}{1 - \cos x} ^ 2$

${f}^{'} = \frac{2 \sin x \cos x - 2 \sin x {\cos}^{2} x - {\sin}^{3} x}{1 - \cos x} ^ 2$

You can write this as

${f}^{'} = \frac{- \sin x \left(- 2 \cos x + 2 {\cos}^{2} x + {\sin}^{2} x\right)}{1 - \cos x} ^ 2$

At this point, you could actually use the same identity to write

${f}^{'} = \frac{- \sin x \cdot \left(- 2 \cos x + {\cos}^{2} x + {\overbrace{{\cos}^{2} x + {\sin}^{2} x}}^{\textcolor{red}{\text{=1}}}\right)}{1 - \cos x} ^ 2$

${f}^{'} = \frac{- \sin x \cdot \left(1 - 2 \cos x + {\cos}^{2} x\right)}{1 - \cos x} ^ 2$

Since you know that

$\textcolor{b l u e}{{\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}}$

you can write

$1 - 2 \cos x + {\cos}^{2} x = {\left(1 - \cos x\right)}^{2}$

Finally, plug this back into the expression for ${f}^{'}$ to get

${f}^{'} = \frac{- \sin x \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(1 - \cos x\right)}^{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(1 - \cos x\right)}^{2}}}}} = \textcolor{g r e e n}{- \sin x}$

So there you have it, the same result, but significantly more work to do.