If $y=e^{\sin ^{-1} x}$, then $\left(1-x^2\right) y_2-x y_1=$

Question

Aicharya · Accepted Answer

Given, $y=e^{\sin ^{-1} x}$ Differentiating w.r. to $x$ $ \begin{aligned} & y_1=\frac{d}{d x} e^{\sin ^{-1} x} \\ & y_1=e^{\sin ^{-1} x} \cdot \frac{d}{d x} \sin ^{-1} x \\ & y_1=e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}} \end{aligned} $

Again differentiating w.r. to $x$, $ \begin{array}{rlrl} & \frac{d}{d x}\left(y_1 \cdot \sqrt{1-x^2}\right) & =\frac{d}{d x}(y) \\ \Rightarrow y_2 \cdot \sqrt{1-x^2}+y_1 \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) & =y_1 \\ \Rightarrow \quad & y_2 \cdot \sqrt{1-x^2}-y_1 \cdot \frac{x}{\sqrt{1-x^2}} & =y_1 \\ \Rightarrow & \frac{y_2\left(1-x^2\right)-x y_1}{\sqrt{1-x^2}} & =y_1 \\ \Rightarrow & y_2\left(1-x^2\right)-x y_1 & =y_1 \sqrt{1-x^2} \\ y_2\left(1-x^2\right)-x y_1 & =y \end{array} $ $\therefore$ Hence, option (c) is correct

If $y=e^{\sin ^{-1} x}$, then $\left(1-x^2\right) y_2-x y_1=$

Solution