If x=a sin^3 θ, y=b cos^3 θ, then find (d^2 y)/(d x^2) at θ=(π)/4.

$$\frac{d x}{d \theta}=3 a \sin ^{2} \theta \cos \theta, \frac{d y}{d \theta}=-3 b \cos ^{2} \theta \sin \theta $$ $\displaystyle \Rightarrow \frac{d y}{d x}=\frac{-3 b \cos ^{2} \theta \sin \theta}{3 a \sin ^{2} \theta \cos \theta}=-\frac{b}{a} \cot \theta$ $\displaystyle \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{b}{a} \operatorname{cosec}^{2} \theta \frac{d \theta}{d x}=\frac{b}{a} \operatorname{cosec}^{2} \theta \cdot \frac{1}{3 a \sin ^{2} \theta \cos \theta}=\frac{b}{3 a^{2}} \sec \theta \operatorname{cosec}^{4} \theta$ $\displaystyle \left.\Rightarrow \frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}=\frac{4 \sqrt{2} b}{3 a^{2}}$