How do you find the derivatives of $x=ln(xy)$ ?

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Answer 1 · 2017-01-07T21:45:05

$(dy)/(dx) = e^x((x-1)/(x^2))$

Take the exponential of both sides of the equation:

$x=ln(xy) => e^x = e^(ln(xy)) = xy$

So $y(x)$ can be made explicit:

$y(x) = e^x/x$

and

$(dy)/(dx) = (xe^x-e^x)/(x^2)= e^x((x-1)/(x^2))$

Answer 2 · 2017-01-07T21:46:23

Use the properties of logarithms and its inverse to write the given equation as a function of y and then use the quotient rule:

Given: $x = ln(xy)$

$x = ln(x) + ln(y)$

$ln(y) = x - ln(x)$

$ln(y) = x + ln(1/x)$

$e^ln(y) = e^(x + ln(1/x))$

$y = e^x/x$

$dy/dx = (x - 1)/x^2e^x$

How do you find the derivatives of x=ln(xy)x=ln(xy)?