Question

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

• A. $u$
• B. $2u$
• C. $3u$
• D. $-3u$

Answer 1

Answer: (C)

$3u$

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

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

$\Rightarrow \frac{\partial u}{\partial x}=y^2\left[f\right(\frac{x}{y})+\frac{1}{y}.f^{\prime }(\frac{x}{y}).x]$

$\Rightarrow \frac{\partial u}{\partial x}=y^{2}f\left(\frac{x}{y}\right)+xy.f^{\prime }\left(\frac{x}{y}\right)$

$\Rightarrow \frac{\partial u}{\partial y}=x\left[2yf\right(\frac{x}{y})-\frac{x}{y^2}.f^{\prime }(\frac{x}{y}).y^2]$

$\Rightarrow \frac{\partial u}{\partial y}=2xyf\left(\frac{x}{y}\right)-x^2{f}^{\prime }\left(\frac{x}{y}\right)$

Now we make calculation as follows,

$\Rightarrow x\frac{\partial u}{\partial x}=x{y}^{2}f\left(\frac{x}{y}\right)+x^{2}y.f^{\prime }\left(\frac{x}{y}\right)$

$\Rightarrow y\frac{\partial u}{\partial y}=2x{y}^{2}f\left(\frac{x}{y}\right)-x^{2}y{f}^{\prime }\left(\frac{x}{y}\right)$

$\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=x{y}^{2}f\left(\frac{x}{y}\right)+x^{2}y.f^{\prime }\left(\frac{x}{y}\right)+2x{y}^{2}f\left(\frac{x}{y}\right)-x^{2}y{f}^{\prime }\left(\frac{x}{y}\right)$

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

$\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=3u$