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

(iii) $\frac{d}{d x} \sin x=-\cos x$

(iv) $\frac{d}{d x} x^{n}=n x^{n-1}$

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

Given: -

$Y=\log (1+\cos x)$

Differentiating w.r.t $x$

$\frac{d y}{d x}=\frac{1}{1+\cos x} \cdot(-\sin x)$

$\Rightarrow \frac{d y}{d x}=\frac{-\sin x}{1+\cos x}$

Differentiating w.r.t.x

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

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

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

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{1+\cos \mathrm{x}}$

Differentiating w.r.t $\mathrm{x}$

$\frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}=-\left(\frac{1}{(1+\cos \mathrm{x})^{2}} \times-\sin \mathrm{x}\right)$

$\Rightarrow \frac{d^{3} y}{d x^{3}}=-\left(\frac{-\sin x}{1+\cos x}\right) \times\left(\frac{-1}{1+\cos x}\right)$

$\Rightarrow \frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}=-\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

$\Rightarrow \frac{\mathrm{d}^{3} \mathrm{y}}{\mathrm{dx}^{3}}+\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=0$