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If $y=\sin \left(\frac{1+x^2}{1-x^2}\right)$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{4 x}{\left(1-x^2\right)^2} \cdot \cos \left(\frac{1+x^2}{1-x^2}\right)$
The given equation is $y=\sin \left(\frac{1+x^2}{1-x^2}\right)$.
Differentiate both sides of the equation with respect to $x$.
$\begin{aligned} & \therefore \frac{\mathrm{d}(y)}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left[\sin \left(\frac{1+x^2}{1-x^2}\right)\right] \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{1+x^2}{1-x^2}\right] \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) .\frac{\left(1-x^2\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(1+x^2\right)-\left(1+x^2\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(1-x^2\right)}{\left(1-x^2\right)^2}\end{aligned}$
$\begin{aligned} {\left[\because \frac{\mathrm{d}}{\mathrm{d x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \cdot \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^2}\right]}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{\left(1-x^2\right)(2 x)-\left(1+x^2\right)(-2 x)}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{2 x-2 x^3+2 x+2 x^3}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{4 x}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right) \\ & \text { Therefore, } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right)\end{aligned}$
Differentiate both sides of the equation with respect to $x$.
$\begin{aligned} & \therefore \frac{\mathrm{d}(y)}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left[\sin \left(\frac{1+x^2}{1-x^2}\right)\right] \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{1+x^2}{1-x^2}\right] \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) .\frac{\left(1-x^2\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(1+x^2\right)-\left(1+x^2\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(1-x^2\right)}{\left(1-x^2\right)^2}\end{aligned}$
$\begin{aligned} {\left[\because \frac{\mathrm{d}}{\mathrm{d x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \cdot \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^2}\right]}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{\left(1-x^2\right)(2 x)-\left(1+x^2\right)(-2 x)}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{2 x-2 x^3+2 x+2 x^3}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{4 x}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right) \\ & \text { Therefore, } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right)\end{aligned}$
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