Question

please find my answers need urgently​

Answer 1

Step by step solution :

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

Differentiating both sides with respect to x, we get

$\mathsf{\frac{dy}{dx}=\frac{d}{dx}\bigg[sin\left(\frac{1+x^{2}}{1-x^{2}}\right)\bigg]}$

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

$\small{\mathsf{=cos\left(\frac{1+x^{2}}{1-x^{2}}\right)\:\frac{(1-x^{2})\frac{d}{dx}(1+x^{2})-(1+x^{2})\frac{d}{dx}(1-x^{2})}{(1-x^{2})^{2}}}}$

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

$\mathsf{=cos\left(\frac{1+x^{2}}{1-x^{2}}\right)\:\frac{2x-2x^{3}+2x+2x^{2}}{(1-x^{2})^{3}}}$

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

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

$\to \boxed{\mathsf{\frac{dy}{dx}=\frac{4x}{(1-x^{2})^{2}}\:cos\left(\frac{1+x^{2}}{1-x^{2}}\right)}}$