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# Find the value

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, When $\quad \mathrm{y}=\sin \left(\frac{1+\mathrm{x}^{2}}{1-\mathrm{x}^{2}}\right)$

Solution:

Let $y=\sin \left(\frac{1+x^{2}}{1-x^{2}}\right), u=1+x^{2}, v=1-x^{2}, z=\frac{1+x^{2}}{1-x^{2}}$

Formula : $\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}=2 \mathrm{x}$ and $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$

According to the quotient rule of differentiation

If $z=\frac{u}{v}$

$\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$

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

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

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

According to the chain rule of differentiation

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}} \times \frac{\mathrm{dz}}{\mathrm{dx}}$

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