Question

$I$f $u=\frac{x^{1/4}+y^{1/4}}{x^{1/5}+y^{1/5}}$,

then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=?$

• A. $u$
• B. $\frac{1}{4}u$
• C. $\frac{1}{5}u$
• D. $\frac{1}{20}u$

Answer 1

The correct option is D $\frac{1}{20}u$

$\frac{\partial u}{\partial x}=\frac{\frac{1}{4}{x}^{-3/4}(x^{1/5}+y^{1/5})-\frac{1}{5}{x}^{-4/5}(x^{1/4}+y^{1/4})}{{(x^{1/5}+y^{1/5})}^2}$
$x\frac{\partial u}{\partial x}=\frac{\frac{1}{4}{x}^{1/4}(x^{1/5}+y^{1/5})-\frac{1}{5}{x}^{1/5}(x^{1/4}+y^{1/4})}{{(x^{1/5}+y^{1/5})}^2}$
$y\frac{\partial u}{\partial y}=\frac{\frac{1}{4}{y}^{1/4}(x^{1/5}+y^{1/5})-\frac{1}{5}{x}^{1/5}(x^{1/4}+y^{1/4})}{{(x^{1/5}+y^{1/5})}^2}$
$\sum x\frac{\partial u}{\partial x}=\frac{1}{20}u$