Differentiation Question 86
Question: If $ x^{x}y^{y}z^{z}=c $ , then $ \frac{\partial z}{\partial x}= $
[EAMCET 1999]
Options:
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
Show Answer
Answer:
Correct Answer: B
Solution:
$ x^{x}y^{y}z^{z}=c $
Therefore $ \log (x^{x}y^{y}z^{z})=\log c $
Therefore $ x\log x+y\log y+z\log z=\log c $
…..(i) Here x, y are regarded as independent variables and z depends on x, y. Differentiating (i) partially w.r.t. -x- $ x.\frac{1}{x}+\log x.1+0+( z.\frac{1}{z}+\log z.1 )\frac{\partial z}{\partial x}=0 $
$ \therefore $ $ \frac{\partial z}{\partial x}=-\frac{1+\log x}{1+\log z} $ .
BETA
