If the solve the problem
Question:

If $x=a \sin t$ and $y=a\left(\cos t+\log \tan \frac{t}{2}\right)$, find $\frac{d^{2} y}{d x^{2}}$

Solution:

Formula: –

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$

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

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

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

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

(vi) parameteric forms $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given: –

$x=$ atsint and $y=a\left(\cos t+\log \tan \left(\frac{t}{2}\right)\right)$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{acost}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\mathrm{asint}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{asint}+\frac{\mathrm{a}}{\tan \left(\frac{\mathrm{t}}{2}\right)} \times \sec ^{2} \frac{\mathrm{t}}{2} \times \frac{1}{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{asint}+\frac{\mathrm{a}}{2 \sin \left(\frac{\mathrm{t}}{2}\right) \cos \left(\frac{\mathrm{t}}{2}\right)}$

$\Rightarrow \frac{d y}{d t}=-a \sin t+a \operatorname{cosect}$

$\Rightarrow \frac{d^{2} y}{d t^{2}}=-a \cos t-a \operatorname{cosectcott}$

$\frac{d^{2} y}{d x^{2}}=\frac{\frac{d x}{d t} \frac{d^{2} y}{d t^{2}}-\frac{d y}{d t} \frac{d^{2} x}{d t^{2}}}{\left(\frac{d x}{d t}\right)^{3}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{acost}(-\mathrm{acost}-\mathrm{acosectcott})-(-\mathrm{asint}+\mathrm{acosect})(-\mathrm{asint})}{(\mathrm{acost})^{3}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-\mathrm{a}^{2}\left(\cos ^{2} \mathrm{t}+\sin ^{2} \mathrm{t}\right)-\mathrm{a}^{2} \cot ^{2} \mathrm{t}+\mathrm{a}^{2}}{\mathrm{a}^{3} \cos ^{3} \mathrm{t}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{\mathrm{a} \sin ^{2} \mathrm{t} \cos \mathrm{t}}$

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