Question
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\text{b}\sin^2\theta\text{ and y}=\text{a}\cos^2\theta$
$\text{x}=\text{b}\sin^2\theta\text{ and y}=\text{a}\cos^2\theta$
Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\text{b}\sin^2\theta)=2\text{b}\sin\theta\cos\theta$
$\frac{\text{dy}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}(\text{a}\cos^2\theta)=2\text{b}\cos\theta\sin\theta$
$\therefore\frac{\text{dy}}{\text{d}\text{x}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-2\text{a}\cos\theta\sin\theta}{2\text{b}\sin\theta\cos\theta}=\frac{-\text{a}}{\text{b}}$
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$\text{f(x)}=\sin^{-1}\text{x}+\sin^{-1}2\text{x}$