If x^x y^y z^z = c, then z x = - Vidyadip

MCQ

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

A
${{1 + \log x} \over {1 + \log z}}$
✓
$ - {{1 + \log x} \over {1 + \log z}}$
C
$ - {{1 + \log y} \over {1 + \log z}}$
D
None of these

✓

Answer

Correct option: B.

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

b
(b) ${x^x}{y^y}{z^z} = c$ ==> $\log ({x^x}{y^y}{z^z}) = \log c$

==> $x\log x + y\log y + z\log z = \log c$ .....$(i)$

Here $x, y$ are regarded as independent variables and $z $ depends on $x, y.$

Differentiating $ (i) $ partially $w.r.t. ‘x’$

$x.\frac{1}{x} + \log x.1 + 0 + \left( {z.\frac{1}{z} + \log z.1} \right)\frac{{\partial z}}{{\partial x}} = 0$

$\therefore $ $\frac{{\partial z}}{{\partial x}} = - \frac{{1 + \log x}}{{1 + \log z}}$.

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