If `y=sqrt{a^{2}-x^{2}}`, prove that `y frac{d y}{d x}+x=0`

Explanation

Given `y=sqrt{a^{2}-x^{2}}`


On differentiating y with respect to `x`, we get


`frac{d y}{d x}=frac{d}{d x}(sqrt{a^{2}-x^{2}})` `Rightarrow frac{d y}{d x}=frac{d}{d x}[(a^{2}-x^{2})^{frac{1}{2}}]` We know `frac{d}{d x}(x^{n})=n x^{n-1}` `Rightarrow frac{d y}{d x}=frac{1}{2}(a^{2}-x^{2})^{frac{1}{2}-1} frac{d}{d x}(a^{2}-x^{2})`


[using chain rule]



`Rightarrow frac{d y}{d x}=frac{1}{2}(a^{2}-x^{2})^{-frac{1}{2}}[frac{d}{d x}(a^{2})-frac{d}{d x}(x^{2})] `


`Rightarrow frac{d y}{d x}=frac{1}{2 sqrt{a^{2}-x^{2}}}[frac{d}{d x}(a^{2})-frac{d}{d x}(x^{2})]`



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