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Question

(D) Given $u = \sin^{-1}\left(\frac{y}{x}\right)$.To find $\frac{\partial u}{\partial x}$,we differentiate $u$ with respect to $x$ while treating $y$

Vedclass · Accepted Answer

$-\frac{y}{x\sqrt{x^2 - y^2}}$

Vedclass · Answer

(D) Given $u = \sin^{-1}\left(\frac{y}{x}\right)$.To find $\frac{\partial u}{\partial x}$,we differentiate $u$ with respect to $x$ while treating $y$ as a constant.Using the chain rule,$\frac{d}{dx}(\sin^{-1}(v)) = \frac{1}{\sqrt{1 - v^2}} \cdot \frac{dv}{dx}$.Here,$v = \frac{y}{x}$,so $\frac{\partial v}{\partial x} = y \cdot \frac{d}{dx}(x^{-1}) = y \cdot (-x^{-2}) = -\frac{y}{x^2}$.Substituting these into the derivative formula:$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - (\frac{y}{x})^2}} \cdot \left(-\frac{y}{x^2}\right)$$= \frac{1}{\sqrt{\frac{x^2 - y^2}{x^2}}} \cdot \left(-\frac{y}{x^2}\right)$$= \frac{x}{\sqrt{x^2 - y^2}} \cdot \left(-\frac{y}{x^2}\right)$$= -\frac{y}{x\sqrt{x^2 - y^2}}$.

If $u = \sin^{-1}\left(\frac{y}{x}\right)$,then $\frac{\partial u}{\partial x}$ is equal to

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