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Find the value

Question:

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

 

Solution:

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

Formula :

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

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

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

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

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

 

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