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How do you find the derivative of y in the equation #ln(xy)=x+y#?

Answer 1

You have to remember that in this "implicit" function #y# is a function of #x# so deriving you get: #1/(xy)*(y+xdy/dx)=1+dy/dx# where you used the Chain Rule on the #ln# Then: #1/x+1/ydy/dx=1+dy/dx# So you collect #dy/dx# and get: #dy/dx=(1-1/x)/(1/y-1)=((x-1)/x)(y/(1-y))=(y(x-1))/(x(1-y)#

Hope it helps

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Answer 2

Luke Alvarez

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

Start by differentiating both sides of the equation with respect to ( x ):

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

Using the chain rule and product rule on the left side:

[ \frac{1}{xy}\frac{d}{dx}(xy) = 1 + \frac{dy}{dx} ]

[ \frac{y + x\frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx} ]

Now, solve for ( \frac{dy}{dx} ):

[ y + x\frac{dy}{dx} = xy + y\frac{dy}{dx} ]

[ x\frac{dy}{dx} - y\frac{dy}{dx} = xy - y ]

[ \frac{dy}{dx}(x - y) = xy - y ]

[ \frac{dy}{dx} = \frac{xy - y}{x - y} ]

So, the derivative of ( y ) with respect to ( x ) is ( \frac{xy - y}{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.