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If $y=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a},$ then $\frac{dy}{dx}=$ |
$\sqrt{a^2-x^2}$ $\sqrt{a^2+x^2}$ $\frac{1}{\sqrt{a^2-x^2}}$ $\frac{1}{\sqrt{a^2+x^2}}$ |
$\sqrt{a^2-x^2}$ |
The correct answer is Option (1) → $\sqrt{a^2-x^2}$ $\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}$ [Integration formula] $\frac{dy}{dx}=\sqrt{a^2-x^2}$ |
