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

Answer 1

Ezekiel Alexander

#(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))#

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Answer 2

Elijah Abram

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#

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Answer 3

Emily Ammerman

To find the derivative of ( x = \ln(xy) ) with respect to ( x ), you can use implicit differentiation.

[ \frac{d}{dx}(x) = \frac{d}{dx}(\ln(xy)) ]

By applying the chain rule and the product rule, you'll get:

[ 1 = \frac{1}{xy}(y + xy') ]

Then, solve for ( y' ):

[ y' = \frac{y}{x} - \frac{1}{x} ]

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Answer from HIX Tutor

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