Question

If $f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1$, then $\Sigma x\frac{\partial f}{\partial x}$ is equal to

• A. $2f$
• B. $f+2$
• C. $2f+2$
• D. $2f-2$

Answer 1

Answer: (C)

$2f+2$

The given information is:

$f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1$

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

$\Rightarrow \frac{\partial f}{\partial x}=\frac{2x}{a^2}$

$\Rightarrow \frac{\partial f}{\partial y}=\frac{2y}{b^2}$

$\Rightarrow \frac{\partial f}{\partial z}=\frac{2z}{c^2}$

$\Rightarrow x\frac{\partial f}{\partial x}=\frac{2x^2}{a^2}$

$\Rightarrow y\frac{\partial f}{\partial y}=\frac{2y^2}{b^2}$

$\Rightarrow z\frac{\partial f}{\partial z}=\frac{2z^2}{c^2}$

$\Rightarrow x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}=\frac{2x^2}{a^2}+\frac{2y^2}{b^2}+\frac{2z^2}{c^2}$

$\Rightarrow x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}=2(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2})$

$\Rightarrow x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}=2(1+f)$

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