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

Answer 1

Emily Ammerman

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

Here's another method.

#y = ln(x+ y)#

#e^y = e^ln(x + y)#

#e^y = x + y#

The derivative of #e^a# is #e^a#. Therefore:

#e^y(dy/dx) = 1 + dy/dx#

#e^y(dy/dx) - dy/dx = 1#

#dy/dx(e^y - 1) = 1#

#dy/dx= 1/(e^y - 1)#

#dy/dx= 1/(e^ln(x+ y) - 1)#

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

Hopefully this helps!

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Answer 2

Aurora Alvarado

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

Differentiate both sides of the equation with respect to ( x ).
Apply the chain rule and the derivative of natural logarithm.
Solve for ( \frac{dy}{dx} ).

The result is:

[ \frac{dy}{dx} = \frac{1}{1 + \frac{dy}{dx}} ]

Rearranging, you get:

[ \frac{dy}{dx} = \frac{1}{1 - \frac{dx}{dy}} ]

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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.