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If $x^y\; y^x = 100$, evaluate $\large \frac{dy}{dx}$

$\begin{array}{1 1}(A) \large \bigg (\Large \frac{\Large \frac{x}{y} + \log y}{\Large \frac{y}{x} + \log x} \bigg) \\ (B) \large \bigg (\Large \frac{\Large \frac{y}{x} + \log x}{\Large \frac{x}{y} + \log y} \bigg) \\ (C) \large \bigg (\Large \frac{\Large \frac{x}{y} + \log x}{\Large \frac{y}{x} + \log y} \bigg) \\(D) \large \bigg (\Large \frac{\Large \frac{y}{x} + \log y}{\Large \frac{x}{y} + \log x} \bigg) \end{array} $

1 Answer

Toolbox:
  • Chain Rule: $d (u, v) = u\; dv + v \; du$
  • $\log x^y \; y^x = x \log y + y \log x$
Given $x^y\; y^x = 100$
Taking $\log$ on both sides, $\log (x^y\; y^x) = \log 100$
$\Rightarrow \log x^y \; y^x = x \log y + y \log x = \log 100$
Differentiating, we get:
$y \; \large\frac{1}{x} $$ + \log x \large \frac{dy}{dx}$$+\large\frac{x}{y}\;\frac{dy}{dx}$$+\log y$ $=0$
Resolving, we get $\large\frac{dy}{dx} $$= \large \bigg (\Large \frac{\Large \frac{y}{x} + \log y}{\Large \frac{x}{y} + \log x} \bigg)$
answered Dec 22, 2013 by balaji.thirumalai
 

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