If \(x=3 \tan \mathrm{t}\) and \(y=3 \sec \mathrm{t}\), then the value of \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) at \(\mathrm{t}=\frac{\pi}{4}\) is

\(\begin{aligned} & x=3 \tan \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=3 \sec ^2 \mathrm{t} \\ & y=3 \sec \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{dt}}=3 \sec \mathrm{t} \tan \mathrm{t} \\ & \text { Now, } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{3 \sec \mathrm{tan} t}{3 \sec ^2 \mathrm{t}}=\sin \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{d}}{\mathrm{dt}}(\sin \mathrm{t}) \cdot \frac{\mathrm{dt}}{\mathrm{d} x} \\ & =\cos t \times \frac{1}{3 \sec ^2 t} \\ & \therefore \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\cos ^3 \mathrm{t}}{3} \\ & \therefore \quad\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)_{\left(\mathrm{t}=\frac{\pi}{4}\right)}=\frac{\left(\cos \frac{\pi}{4}\right)^3}{3}=\frac{1}{6 \sqrt{2}} \\ & \end{aligned}\)