If $$Sin-1x2-y2x2+y2=log$$ a then $$dydx=$$

The given equation is $$Sin-1x2-y2x2+y2=log$$ ait implies that $$x2-y2x2+y2=sinlogaDifferentiate$$ both sides with respect to $$xx2+y22x-2ydydx-x2-y22x+2ydydxx2+y22=0dydx2yx2+y2+2yyx2-y2=2xx2+y2-2xx2-y2dydx=2xy22x2y$$ $$=yxTherefore$$, $$dydx=yx$$

Questions

If ${Sin}^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a$ then $\frac{dy}{dx} =$

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\frac{x}{y}

\frac{y}{x}

\frac{2 y^{2}}{{(x^{2} + y^{2})}^{2}}

\frac{2 x^{2}}{(x^{2} + y^{2})}

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detailed solution

Correct option is B

The given equation is ${Sin}^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a$

it implies that $\frac{x^{2} - y^{2}}{x^{2} + y^{2}} = \sin (\log a)$

Differentiate both sides with respect to $x$

$\frac{(x^{2} + y^{2}) (2 x - 2 y \frac{d y}{d x}) - (x^{2} - y^{2}) (2 x + 2 y \frac{d y}{d x})}{{(x^{2} + y^{2})}^{2}} = 0$

$\begin{array}{rcl} \frac{d y}{d x} (2 y (x^{2} + y^{2}) + 2 y y (x^{2} - y^{2})) & = & 2 x (x^{2} + y^{2}) - 2 x (x^{2} - y^{2}) \\ \frac{d y}{d x} & = & \frac{2 x y^{2}}{2 x^{2} y} \\ = & \frac{y}{x} \end{array}$