Question

If y = cos⁻¹ x, find $\frac{d^{2}y}{d{x}^2}$ in terms of y alone.

Answer 1

Here,
$$\begin{gathered} y={\cos }^{-1}x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{-1}{\sqrt{1-x^2}} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{-2x}{2{\sqrt{1-x^2}}^{3/2}}=\frac{-x}{{\left(1-x^2\right)}^{3/2}} \\ \mathrm{Now}, \\ y={\cos }^{-1}x \\ \Rightarrow x=\cos y \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{-\cos y}{{\left(1-{\cos }^{2}y\right)}^{3/2}}=-\frac{\cos y}{{\left({\sin }^{2}y\right)}^{3/2}}=-\cot y{\mathrm{cosec}}^{2}y \end{gathered}$$