Question

U=tan^-1(xy/√1+x^2+y^2) ka partial differentiation

Answer 1

U=tan^-1(xy/√1+x^2+y^2) ka partial differentiation

Given,

U=tan^-1(xy/√1+x^2+y^2)

$\dfrac{\partial \:}{\partial \:x}\left(\arctan \left(\dfrac{xy}{\sqrt{1+x^2+y^2}}\right)\right)$

Apply chain rule,

$f=\arctan \left(u\right),\:\:u=\dfrac{xy}{\sqrt{1+x^2+y^2}}$

$=\dfrac{\partial \:}{\partial \:u}\left(\arctan \left(u\right)\right)\dfrac{\partial \:}{\partial \:x}\left(\dfrac{xy}{\sqrt{1+x^2+y^2}}\right)$

$=\dfrac{1}{u^2+1}\cdot \dfrac{y\left(y^2+1\right)}{\left(1+x^2+y^2\right)\sqrt{1+x^2+y^2}}$

$=\dfrac{1}{\left(\frac{xy}{\sqrt{1+x^2+y^2}}\right)^2+1}\cdot \dfrac{y\left(y^2+1\right)}{\left(1+x^2+y^2\right)\sqrt{1+x^2+y^2}}$

$=\dfrac{x^2+y^2+1}{x^2y^2+x^2+y^2+1}\cdot \dfrac{y\left(y^2+1\right)}{\left(x^2+y^2+1\right)\sqrt{x^2+y^2+1}}$

$=\dfrac{\left(1+x^2+y^2\right)y\left(y^2+1\right)}{\left(x^2y^2+1+x^2+y^2\right)\left(1+x^2+y^2\right)\sqrt{1+x^2+y^2}}$

$=\dfrac{y\left(y^2+1\right)}{\left(x^2y^2+1+x^2+y^2\right)\sqrt{1+x^2+y^2}}$