# How do you find the second derivative of sin(x)/(1-cos(x))?

Aug 14, 2015

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

#### Explanation:

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

Aug 15, 2015

You take the first derivative twice.

$\textcolor{g r e e n}{{d}^{2} / \left({\mathrm{dx}}^{2}\right) \left[f \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left[\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]\right]}$

$= \frac{d}{\mathrm{dx}} \left[\frac{\left(1 - \cos x\right) \left(\cos x\right) - \sin x \left(0 + \sin x\right)}{1 - \cos x} ^ 2\right]$

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

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

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

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

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

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