Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
If $\text{y}=\sqrt{\text{a}^2-\text{x}^2},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
✓
Answer
DIfferentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{a}^2-\text{x}^2}\big)$
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}\frac{\text{d}}{\text{dx}}\big(\text{a}^2-\text{x}^2\big)$
[Using chain rule]
$=\frac{1}{2\sqrt{\text{a}^2-\text{x}^2}}(-2\text{x})$
$=\frac{-\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}$
$\big[\text{Since},\sqrt{\text{a}^2-\text{x}^2}=\text{y}\big]$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}$
Hence, the solution is, $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=0$
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