How do you differentiate sin(x)/(1-cos(x))?

Jul 29, 2015

You use the quotient rule.

Explanation:

Every time you're dealing with a function that is actually the quotient of two other functions, let's call them $f \left(x\right)$ and $g \left(x\right)$, you can differentiate said function by using the quotient rule.

For $y = f \frac{x}{g} \left(x\right)$, you have

color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2, where $g \left(x\right) \ne 0$

In your case, you have

$y = \sin \frac{x}{1 - \cos x}$, with

$\left\{\begin{matrix}f \left(x\right) = \sin x \\ g \left(x\right) = 1 - \cos x\end{matrix}\right.$

You alsoe need to remember that

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

and that

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

So, the derivatives of $f \left(x\right)$ and $g \left(x\right)$ will be

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

$\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(1 - \cos x\right)$

$\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(1\right) - \frac{d}{\mathrm{dx}} \left(\cos x\right) = 0 - \left(- \sin x\right) = \sin x$

The derivative of your function $y$ will thus be

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

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

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

You can simplify this further by using the fact that

color(blue)(sin^2x + cos^2x = 1

to get

${y}^{'} = \textcolor{g r e e n}{\frac{\cos x - 1}{1 - \cos x} ^ 2}$