Question

If $u={\sin }^{-1}\left[\right(x^2+y^2{)}^{\frac{1}{5}}]$ then $\sum x\frac{\partial u}{\partial x}=$

• A. $\frac{2}{5}\cot u$
• B. $\frac{5}{2}\cot u$
• C. $\frac{2}{5}\tan u$
• D. $\frac{5}{2}\tan u$

Answer 1

Answer: (C)

$\frac{2}{5}\tan u$

$u={\sin }^{-1}\left[\right(x^2+y^2{)}^{\frac{1}{5}}]$

$\sin u={(x^2+y^2)}^{\frac{1}{5}}$ ..... $\left(i\right)$

Differentiating $\left(i\right)$ partially w.r.t $x$, we get

$\cos u\frac{\partial u}{\partial x}=\frac{1}{5}\frac{2x}{(x^2+y^2{)}^{\frac{4}{5}}}$..... $\left(ii\right)$

Differentiating $\left(i\right)$ partially w.r.t $y$, we get

$\cos u\frac{\partial u}{\partial y}=\frac{2}{5}\frac{y}{(x^2+y^2{)}^{\frac{4}{5}}}$..... $\left(iii\right)$

From $\left(ii\right)$ and $\left(iii\right)$, we get

$\cos u(\frac{x\partial u}{\partial x}+\frac{y\partial u}{\partial y})=\frac{2}{5}\sin u$

$\Rightarrow \frac{x\partial u}{\partial x}+\frac{y\partial u}{\partial y}=\frac{2}{5}\tan u$