Question

If $y=\sin \left(\frac{1+x^2}{1-x^2}\right)$, then $\frac{dy}{dx}=$

• A. $\frac{4x}{1-x^2}\cos \left(\frac{1+x^2}{1-x^2}\right)$
• B. $\frac{x}{1-x^2}\cos \left(\frac{1+x^2}{1-x^2}\right)$
• C. $\frac{4x}{{\left(1-x^2\right)}^2}\cos \left(\frac{1+x^2}{1-x^2}\right)$
• D. $\frac{x}{{\left(1-x^2\right)}^2}\cos \left(\frac{1+x^2}{1-x^2}\right)$

Answer 1

The correct option is C

$\frac{4x}{{\left(1-x^2\right)}^2}\cos \left(\frac{1+x^2}{1-x^2}\right)$

The explanation for the correct option:

The given equation is $y=\sin \left(\frac{1+x^2}{1-x^2}\right)$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \therefore \frac{d\left(y\right)}{dx}=\frac{d}{dx}\left[\sin \left(\frac{1+x^2}{1-x^2}\right)\right] \\ \Rightarrow \frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)·\frac{d}{dx}\left[\frac{1+x^2}{1-x^2}\right] \\ \Rightarrow \frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)·\frac{\left(1-x^2\right)·\frac{d}{dx}\left(1+x^2\right)-\left(1+x^2\right)·\frac{d}{dx}\left(1-x^2\right)}{{\left(1-x^2\right)}^2}\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{v}\mathbf{·}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathbf{u}\mathbf{·}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}}{{\mathbf{v}}^{\mathbf{2}}}\right] \\ \Rightarrow \frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)·\frac{\left(1-x^2\right)\left(2x\right)-\left(1+x^2\right)\left(-2x\right)}{{\left(1-x^2\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)·\frac{2x-2x^3+2x+2x^3}{{\left(1-x^2\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)·\frac{4x}{{\left(1-x^2\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{4x}{{\left(1-x^2\right)}^2}\cos \left(\frac{1+x^2}{1-x^2}\right) \end{gathered}$$

Therefore, $\frac{dy}{dx}=\frac{4x}{{\left(1-x^2\right)}^2}\cos \left(\frac{1+x^2}{1-x^2}\right)$.

Hence, (C) is the correct option.