Question

If $x^x{y}^y{z}^z=c$, then $\frac{\partial z}{\partial x}=$

• A. $\frac{1+\log x}{1+\log z}$
• B. $-\frac{1+\log x}{1+\log z}$
• C. $\frac{1+\log y}{1+\log z}$
• D. None of these

Answer 1

The correct option is B

$-\frac{1+\log x}{1+\log z}$

Explanation for the correct option.

$x^x{y}^y{z}^z=c$

By taking log on both sides, we get

$x\log x+y\log y+z\log z=\log c$

By differentiating with respect to $x$, we get

$\begin{array}{rcl}\left[\log x\left(1\right)+x\left(\frac{1}{x}\right)\right]+0+\left[\log z\left(\frac{\partial z}{\partial x}\right)+z\left(\frac{1}{z}\times \frac{\partial z}{\partial x}\right)\right]& =& 0\\ \Rightarrow \log x+1+\frac{\partial z}{\partial x}\left(\log z+1\right)& =& 0\\ \Rightarrow \frac{\partial z}{\partial x}& =& -\frac{1-\log x}{1+\log z}\end{array}$

Hence, option B is correct.