Question

If $u=\log \left(\frac{x^4+y^4}{x^2+y^2}\right)$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ is equal to

• A. $2$
• B. $2u$
• C. $3u$
• D. $u$

Answer 1

The correct option is A $2$

The given information is: $u=\log \left(\frac{x^4+y^4}{x^2+y^2}\right)$

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

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

$\Rightarrow \frac{\partial u}{\partial x}=\frac{4x^5+4x^3{y}^2-2x^5-2x{y}^4}{(x^4+y^4)(x^2+y^2)}$

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

$\Rightarrow \frac{\partial u}{\partial y}=\frac{4y^5+4x^2{y}^3-2y^5-2x^{4}y}{(x^4+y^4)(x^2+y^2)}$

Now we make calculation as follows,

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

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

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

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

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

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