Question

If $\sin u=\frac{x+2y+3z}{\sqrt{x^8+y^8+z^8}}$, then $\Sigma x\frac{\partial u}{\partial x}$ is equal to

• A. $3\sin u$
• B. $-3co\sec u$
• C. $3\tan u$
• D. $-3\tan u$

Answer 1

Answer: (D)

$-3\tan u$

Given, $\sin u=\frac{x+2y+3z}{\sqrt{x^8+y^8+z^8}}$

Differentiating on both sides, we get,

$\cos u\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x+2y+3z}{\sqrt{x^8+y^8+z^8}}\right)\Rightarrow \frac{\partial u}{\partial x}=\frac{\tan u}{\sin u}\frac{\partial }{\partial x}\left(\frac{x+2y+3z}{\sqrt{x^8+y^8+z^8}}\right)\Rightarrow \sum x\frac{\partial u}{\partial x}=\frac{\sqrt{x^8+y^8+z^8}}{x+2y+3z}\tan u\left(\frac{(x^8+y^8+z^8)(1x+2y+3z)-(x+2y+3z)(4x^{7}x+4y^{7}y+4z^{7}z)}{(x^8+y^8+z^8)\sqrt{x^8+y^8+z^8}}\right)\Rightarrow \sum x\frac{\partial u}{\partial x}=\tan u(1-4)\Rightarrow \sum x\frac{\partial u}{\partial x}=-3\tan u$

So, the correct option is D.