Question:
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}

Answer:

y= x.sinx

Differentiating both sides of this equation with respect to x, we get:

\begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}

\begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}

\begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}

Differentiating both sides of this equation with respect to x, we get:

L.H.S. =xy^'= x(sinx + xcosx)

\begin{align} =x.sinx + x^2.cosx\end{align}

\begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}

\begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}

\begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}

\begin{align} =y + x.\sqrt{x^2-y^2}\end{align}

R.H.S.

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Question:
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}

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Question: \begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}

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Question:
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}