Question

$I$: If $f(x,y)=\tan [{\tan }^{-1}x+{\tan }^{-1}y]$, then $\frac{f_x}{f_y}=\frac{1+y^2}{1+x^2}$

$II$: If $u=\frac{(x^2+y^2{)}^2}{\sqrt{x^8+y^8}}$, then $x{u}_x+y{u}_y=-4u$

Which of the above statements is true?

• A. Only $I$
• B. Only $II$
• C. Both $I$ and $II$
• D. Neither $I$ nor $II$

Answer 1

The correct option is A Only $I$

$I.f(x,y)=\frac{x+y}{1-xy}$

$\frac{\partial f}{\partial x}=\frac{(1-xy)+(x-y)y}{(1-xy{)}^2}=\frac{1+y^2}{(1-xy{)}^2}$

$\frac{\partial f}{\partial y}=\frac{(1-xy)+(x-y)x}{(1-xy{)}^2}=\frac{(1+x^2)}{(1-xy{)}^2}$

$\frac{f_x}{f_y}=\frac{1+y^2}{1+x^2}$

$II.\frac{\partial u}{\partial x}=\frac{2(x^2+y^2)\left(2x\right)\sqrt{x^8+y^8}-(x^2+y^2{)}^2\frac{1}{2}\left(\frac{8\times x^7}{\sqrt{x^8+y^8}}\right)}{x^8+y^8}$

$=\frac{4(x^2+y^2)\left(x\right(x^8+y^8)-(x^2+y^2\left)x^7\right)}{(x^8+y^8{)}^{3/2}}$

$=\frac{4(x^2+y^2)x{y}^8-y^2{x}^7}{(x^8+y^8{)}^{3/2}}$

$\frac{\partial u}{\partial y}=\frac{\sqrt{x^8+y^8}2(x^2+y^2)\left(2y\right)-(x^2+y^2{)}^2\left(\frac{1}{2}\frac{8y^7}{\sqrt{x^8+y^8}}\right)}{(x^8+y^8)}$

$=\frac{4(x^2+y^2)(y{x}^8-x^2{y}^7)}{(x^8+y^8{)}^{3/2}}$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac{4(x^2+y^2)(x^2{y}^8-y^2{x}^8+y^2{x}^8-x^2{y}^8)}{(x^8+y^8{)}^{3/2}}$

$=0$

Hence, only statement $I$ is true.