Question

$I$f $u={\tan }^{-1}\left(\frac{3x^{2}y+2x{y}^2}{x+y}\right)$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ is equal to

• A. $0$
• B. $\tan u$
• C. $\sin u$
• D. $\sin 2u$

Answer 1

The correct option is D $\sin 2u$

The given information is: $u={\tan }^{-1}\left(\frac{3x^{2}y+2x{y}^2}{x+y}\right)$

Taking partial differentiation w.r.t. to x and y one at a time we get,

$\Rightarrow \frac{\partial u}{\partial x}=\frac{1}{1+\frac{x^2{y}^2(3x+2y{)}^2}{(x+y{)}^2}}.\frac{(6xy+2y^2)(x+y)-(3x^{2}y+2x{y}^2)}{(x+y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial x}=\frac{(6xy+2y^2)(x+y)-(3x^{2}y+2x{y}^2)}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial x}=\frac{6x^{2}y+6x{y}^2+2x{y}^2+2y^3-3x^{2}y-2x{y}^2}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial x}=\frac{3x^{2}y+6x{y}^2+2y^3}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial y}=\frac{1}{1+\frac{x^2{y}^2(3x+2y{)}^2}{(x+y{)}^2}}.\frac{(6xy+3x^2)(x+y)-(3x^{2}y+2x{y}^2)}{(x+y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial y}=\frac{(6xy+3x^2)(x+y)-(3x^{2}y+2x{y}^2)}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial y}=\frac{3x^3+3x^{2}y+6x^{2}y+6x{y}^2-3x^{2}y-2x{y}^2}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow \frac{\partial u}{\partial y}=\frac{3x^3+6x^{2}y+4x{y}^2}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

Now we make calculation as follows,

$\Rightarrow x\frac{\partial u}{\partial x}=\frac{3x^{3}y+6x^2{y}^2+2x{y}^3}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow y\frac{\partial u}{\partial y}=\frac{3x^{3}y+6x^2{y}^2+4x{y}^3}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

$\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac{6x^{3}y+6x{y}^3+12x^2{y}^2}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

Now, $\tan u=\frac{3x^{2}y+2x{y}^2}{x+y}$

$\Rightarrow \sin u=\frac{3x^{2}y+2x{y}^2}{\sqrt{(x+y{)}^2+(3x^{2}y+2x{y}^2{)}^2}}$

$\Rightarrow \cos u=\frac{x+y}{\sqrt{(x+y{)}^2+(3x^{2}y+2x{y}^2{)}^2}}$

$\Rightarrow \sin 2u=2\sin u\cos u$

$\Rightarrow \sin 2u=\frac{2(x+y)(3x^{2}y+2x{y}^2)}{(x+y{)}^2+(3x^{2}y+2x{y}^2{)}^2}$

$\Rightarrow \sin 2u=\frac{6x^{3}y+6x{y}^3+12x^2{y}^2}{(x+y{)}^2+x^2{y}^2(3x+2y{)}^2}$

So, $\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u$