Calculate dy/dx if Ln(x+y)=e^x/y

This is an add on to the question above. Thought this might help here are the answer choices:

A. e^x/y xy+e^x/y y^2+y^2
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e^x/y xy+e^x/y x^2+y^2

B. e^x/y xy+e^x/y y^2-y^2
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e^x/y xy-e^x/y x^2+y^2

C. e^x/y xy+e^x/y y^2+y^2
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e^x/y xy-e^x/y x^2+y^2

D. e^x/y xy+e^x/y y^2-y^2
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e^x/y xy+e^x/y x^2+y^2

Hope these will help!

just work it with implicit differentiation:

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

1/(x+y) * (1 + y') = e^x/y (1/y - x/y^2 y')

1/(x+y) + 1/(x+y) y' = 1/y e^x/y - e^x/y xy'/y^2

(1/(x+y) + e^x/y x/y^2)y' = 1/y e^x/y - 1/(x+y)

Putting it all over (x+y)y^2 we have

[y(x+y)e^x/y - y^2]/[y^2 + x(x+y)e^x/y]
= D

calculate dy/dx when y=xy3+y=3x.

To calculate dy/dx, we can use implicit differentiation. Here's how:

Step 1: Differentiate both sides of the equation with respect to x.

On the left side, we have Ln(x+y). When differentiating this with respect to x, we need to apply the chain rule. The derivative of Ln(u), where u is a function of x, is given by (1/u) * du/dx.

So, differentiating Ln(x+y) with respect to x, we get:
(1/(x+y)) * (1 + dy/dx)

On the right side, we have e^(x/y). To differentiate this with respect to x, we need to apply the chain rule here as well. The derivative of e^(u), where u is a function of x, is given by e^(u) * du/dx.

So, differentiating e^(x/y) with respect to x, we get:
e^(x/y) * (1/y) * dy/dx

Step 2: Set the two derivatives equal to each other.

We have:
(1/(x+y)) * (1 + dy/dx) = e^(x/y) * (1/y) * dy/dx

Step 3: Solve for dy/dx.

Now, we can isolate dy/dx in this equation. First, let's multiply both sides of the equation by (x+y) * y to get rid of the denominators:

y + (x+y) * (1 + dy/dx) = e^(x/y) * (x+y) * dy/dx

Simplifying this expression, we get:
y + (x+y) + (x+y) * dy/dx = e^(x/y) * (x+y) * dy/dx

Rearranging the terms, we have:
y + x + y + (x+y) * dy/dx - e^(x/y) * (x+y) * dy/dx = 0

Combining like terms, we get:
2x + 2y + [(x+y) - e^(x/y) * (x+y)] * dy/dx = 0

Finally, isolating dy/dx, we get:
dy/dx = -(2x + 2y) / [(x+y) - e^(x/y) * (x+y)]