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Question:

$y=\sqrt{\frac{1-x}{1+x}}$, prove that $\left(1-x^{2}\right) \frac{d y}{d x}+y=0$

 

Solution:

Let $y=\sqrt{\frac{1-x^{1}}{1+x^{1}}}, u=1-x^{1}, v=1+x^{1}, z=\frac{1-x^{1}}{1+x^{1}}$

According to quotient rule of differentiation

If $\mathrm{z}=\frac{\mathrm{u}}{\mathrm{V}}$

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

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

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

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

According to the chain rule of differentiation

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

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

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

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

(Muliplying and dividing by 1-x )

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

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

Therefore

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

$\left(1-x^{2}\right) \frac{d y}{d x}+y=0$

HENCE PROVED

 

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