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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}$
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Answer

We have, $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1$
Let $\text{x}=\sin\text{A},\text{y}=\sin\text{B}$
$\Rightarrow\sin\text{B}\sqrt{1-\sin^2\text{A}}+\sin\text{A}\sqrt{1-\sin^2\text{B}}=1$
$\Rightarrow\sin\text{B}\cos\text{A}+\sin\text{A}\cos\text{B}=1$
$\big[\because\sin(\text{x}+\text{y})=\sin\text{x}\cos\text{y}+\cos\text{x}\sin\text{y}\big]$
$\Rightarrow\sin\big(\text{A}+\text{B}\big)=1$
$\Rightarrow\text{A}+\text{B}=\sin^{-1}(1)$
$\Rightarrow\sin^{-1}\text{x}+\sin^{-1}\text{y}=\frac{\pi}{2} \big[\because\text{x}=\sin\text{A},\text{y}=\sin\text{B}\big]$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sin^{-1}\text{y}\big)=\frac{\text{d}}{\text{dx}}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}+\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\sqrt\frac{1-\text{y}^2}{1-\text{x}^2}$

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