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Solution:

Formula: -

(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$

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

(iii) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wous })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

Given: $-y=\cos ^{-1} x$

Then,

$\frac{d y}{d x}=\frac{d\left(\cos ^{-1} x\right)}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1+x^{2}}}$

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

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-1 \cdot\left(1-\mathrm{x}^{2}\right)^{-\frac{3}{2}}}{2} \cdot \frac{\mathrm{d}\left(1-\mathrm{x}^{2}\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{2 \sqrt{\left(1-\mathrm{x}^{2}\right)^{3}}} \cdot(-2 \mathrm{x})$

$\frac{d^{2} y}{d x^{2}}=\frac{-x}{\sqrt{\left(1-x^{2}\right)^{3}}} \ldots \ldots$ (i)

$y=\cos ^{-1} x$

$\Rightarrow x=\cos y$

Putting $x=\cos y$ in equation(i), we obtain

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\left(1-\cos ^{2} y\right)^{3}}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sqrt{\left(\sin ^{2} y\right)^{3}}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sin ^{3} y}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{\sin y} \cdot \frac{1}{\sin ^{2} y}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\operatorname{coty} \cdot \operatorname{cosec}^{2} y$