Question

If ${\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a\mathrm{then}\frac{dy}{dx}$ is equal to

(a) $\frac{x^2-y^2}{x^2+y^2}$

(b) $\frac{y}{x}$

(c) $\frac{x}{y}$

(d) none of these

Answer 1

(b) $\frac{y}{x}$
$$\begin{gathered} \mathrm{We}\mathrm{have},{\sin }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a \\ \Rightarrow \frac{x^2-y^2}{x^2+y^2}=\sin \log a \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{\left(x^2+y^2\right)\left(2x-2y{\displaystyle \frac{dy}{dx}}\right)-\left(x^2-y^2\right)\left(2x+2y{\displaystyle \frac{dy}{dx}}\right)}{{\left(x^2+y^2\right)}^2}=0 \\ \Rightarrow \frac{2x^3-2x^{2}y{\displaystyle \frac{dy}{dx}}+2x{y}^2-2y^3{\displaystyle \frac{dy}{dx}}-2x^3-2x^{2}y{\displaystyle \frac{dy}{dx}}+2x{y}^2+2y^3{\displaystyle \frac{dy}{dx}}}{{\left(x^2+y^2\right)}^2}=0 \\ \Rightarrow -4x^{2}y\frac{dy}{dx}+4x{y}^2=0 \\ \Rightarrow -4x^{2}y\frac{dy}{dx}=-4x{y}^2 \\ \Rightarrow \frac{dy}{dx}=\frac{4x{y}^2}{4x^{2}y} \\ \therefore \frac{dy}{dx}=\frac{y}{x} \end{gathered}$$