SELECT * FROM question_mgmt as q WHERE id=1156 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=6 AND chapterId=96 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
\begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align}
\begin{align} y= \sqrt{1+x^2}\end{align}
Differentiating both sides of the equation with respect to x, we get:
\begin{align} y^{'}=\frac{d}{dx}\left(\sqrt{1+x^2} \right)\end{align}
\begin{align} y^{'}=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)\end{align}
\begin{align} y^{'}=\frac{2x}{2\sqrt{1+x^2}}\end{align}
\begin{align} y^{'}=\frac{x}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{\sqrt{1+x^2}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{y}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{xy}{1+x^2}\end{align}
∴ L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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