# Differentiate w.r.t x:

Question:

Differentiate w.r.t x:

$\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$

Solution:

Let $y=\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}, u=a^{2}-x^{2}, v=a^{2}+x^{2}, z=\frac{a^{2}-x^{2}}{a^{2}+x^{2}}$

Formula :

$\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}=2 \mathrm{x}$

According to the quotient rule of differentiation

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

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

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

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

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

According to the chain rule of differentiation

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

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

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

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