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How do you implicitly differentiate #ln(xy)=x+y #?

Answer 1

Isabella Ackerman

#(xy-y)/(x-xy)#

Given expression

#ln(xy)=x+y #

#=>lnx+lny=x+y#

Differentiating w.r. to x we can write

#(d(lnx))/(dx)+(d(lny))/(dx)=(d(x))/(dx)+(d(y))/(dx)#

#=>1/x+1/y*(dy)/(dx)=1+(dy)/(dx)#

#=>1/y*(dy)/(dx)-(dy)/(dx)=1-1/x#

#=>(1/y-1)(dy)/(dx)=(x-1)/x#

#=>((1-y)/y)(dy)/(dx)=(x-1)/x#

#=>(dy)/(dx)=(x-1)/x xx(y)/(1-y)=(xy-y)/(x-xy)#

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Answer 2

Christopher Allers

To implicitly differentiate ( \ln(xy) = x + y ):

Start by differentiating both sides of the equation with respect to ( x ).
Use the chain rule when differentiating ( \ln(xy) ).
Apply the product rule when differentiating ( xy ).
Differentiate ( x ) and ( y ) separately.
Solve for ( \frac{dy}{dx} ) after differentiating both sides of the equation.

By following these steps, you can implicitly differentiate ( \ln(xy) = x + y ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.