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

Question

20074 views around the world

You can reuse this answer
Creative Commons License

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#