# Question #d61e2

Feb 27, 2017

Given $y = \csc x$, to find $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$

we know that $\csc x = \frac{1}{\sin} x$
Hence, $y = \frac{1}{\sin} x = {\left(\sin x\right)}^{-} 1$

Differentiating both sides we get
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(\sin x\right)}^{-} 2 \cos x$

Differentiating again and using product rule
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left[- {\left(\sin x\right)}^{-} 2 \cos x\right]$
$\implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \left[{\left(\sin x\right)}^{-} 2 \left(- \sin x\right) + \cos x \left(- 2 {\left(\sin x\right)}^{-} 3 \cos x\right)\right]$
$\implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left[{\left(\sin x\right)}^{-} 1 + 2 {\cos}^{2} x {\left(\sin x\right)}^{-} 3\right]$
$\implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \csc x + 2 {\cot}^{2} x \csc x$
$\implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \csc x \left(1 + 2 {\cot}^{2} x\right)$