If $$y=\sin \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)$$, then $$\frac{dy}{dx}=$$ [AISSE 1987]

Question

If $$y=\sin \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)$$, then $$\frac{dy}{dx}=$$  [AISSE 1987]

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	If \[y=\sin \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\], then \[\frac{dy}{dx}=\] [AISSE 1987]
A.	\[\frac{4x}{1-{{x}^{2}}}.\cos \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\]
B.	\[\frac{x}{{{(1-{{x}^{2}})}^{2}}}.\cos \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\]
C.	\[\frac{x}{(1-{{x}^{2}})}.\cos \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\]
D.	\[\frac{4x}{{{(1-{{x}^{2}})}^{2}}}.\cos \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\]
Answer» E.

If \[y=\sin \left( \frac{1+{{x}^{2}}}{1-{{x}^{2}}} \right)\], then \[\frac{dy}{dx}=\] [AISSE 1987]