Question

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

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

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

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

$y=\sin \left(\frac{1+x^2}{1-x^2}\right)$

Differentiating with respect to $x$ both sides:

$\frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)\times \frac{d}{dx}\left(\frac{1+x^2}{1-x^2}\right)$

Here, we apply Quotient rule of differentiation,

$$\begin{gathered} \frac{dy}{dx}=\cos \left(\frac{1+x^2}{1-x^2}\right)\times \left(\frac{\left(1-x^2\right)\times 2x-\left(1+x^2\right)\times \left(-2x\right)}{{\left(1-x^2\right)}^2}\right)\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{du}{dx}}}{v^2}\right] \\ =\cos \left(\frac{1+x^2}{1-x^2}\right)\times \left(\frac{2x\left(1-x^2+1+x^2\right)}{{\left(1-x^2\right)}^2}\right) \\ =\cos \left(\frac{1+x^2}{1-x^2}\right)\frac{4x}{{\left(1-x^2\right)}^2} \end{gathered}$$

Hence, Option (C) is the correct option.