How do you find the derivative of $x ln y - y ln x = 1$ $x ln y - y ln x = 1$ ?

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How do you find the derivative of $x ln y - y ln x = 1$ $x ln y - y ln x = 1$ ?

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Answer 1 · 2017-06-27T19:29:49

$\frac{d y}{d x} = \frac{\frac{y}{x} - ln y}{\frac{x}{y} - ln x}$ $dy/dx=(y/x-lny)/(x/y-lnx)$

Explanation:

$using implicit differentiation$ $"using "color(blue)"implicit differentiation"$

$differentiate x ln y and y ln x using the$ $"differentiate " xlny" and " ylnx" using the"$
$product rule$ $color(blue)"product rule"$

$(x . \frac{1}{y} . \frac{d y}{d x} + ln y) - (y . \frac{1}{x} + ln x . \frac{d y}{d x}) = 0$ $(x. 1/y . dy/dx+lny)-(y. 1/x+lnx. dy/dx)=0$

$\Rightarrow \frac{x}{y} \frac{d y}{d x} + ln y - \frac{y}{x} - ln x \frac{d y}{d x} = 0$ $rArrx/ydy/dx+lny-y/x-lnxdy/dx=0$

$\Rightarrow \frac{d y}{d x} (\frac{x}{y} - ln x) = \frac{y}{x} - ln y$ $rArrdy/dx(x/y-lnx)=y/x-lny$

$\Rightarrow \frac{d y}{d x} = \frac{\frac{y}{x} - ln y}{\frac{x}{y} - ln x}$ $rArrdy/dx=(y/x-lny)/(x/y-lnx)$

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How do you find the derivative of $x ln y - y ln x = 1$ $x ln y - y ln x = 1$ ?

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