Question

If $x^x.y^y.z^Z=a$ constant, then $\frac{\partial z}{\partial x}$ is equal to

• A. $-\left(\frac{1+{\log }_{e}x}{1+{\log }_{e}z}\right)$
• B. $\left(\frac{1+{\log }_{e}z}{1+{\log }_{e}x}\right)$
• C. $x^y.y^y$
• D. $x^y.y^y.z$

Answer 1

The correct option is A $-\left(\frac{1+{\log }_{e}x}{1+{\log }_{e}z}\right)$

The given information in the question is:

$x^x.y^y.z^z=a$ where $a$ is any constant

Now taking ${\log }_e$ on both sides we get,

$\Rightarrow {\log }_e{x}^x.y^y.z^z={\log }_{e}a$

Using the properties of logarithm that are ${\log }_{e}ab={\log }_{e}a+{\log }_{e}b$ and ${\log }_e{a}^n=n{\log }_{e}a$ we get,

$\Rightarrow {\log }_e{x}^x+{\log }_e{y}^y+{\log }_e{z}^z={\log }_{e}a$

$\Rightarrow x{\log }_{e}x+y{\log }_{e}y+z{\log }_{e}z={\log }_{e}a$

$\Rightarrow z{\log }_{e}z={\log }_{e}a-x{\log }_{e}x-y{\log }_e$

Taking partial differentiation w.r.t to $x$ on both sides we get,

$\Rightarrow \frac{\partial z}{\partial x}(1+{\log }_{e}z)=-(1+{\log }_{e}x)+0$

$\Rightarrow \frac{\partial z}{\partial x}=-\left(\frac{1+{\log }_{e}x}{1+{\log }_{e}z}\right)$