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If xy+7e^(y)=7e, find the second derivative at the point where x=0

User Kleopatra
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Final answer:

If xy+7e^(y)=7e, the second derivative at the point where x=0 is d²y/dy² + e^y = 0.

Step-by-step explanation:

To find the second derivative at the point where x=0, we first need to differentiate the given equation with respect to y. This will give us a differential equation in terms of x and y. Let's differentiate the equation:

d/dy(xy+7e^y) = d/dy(7e)

Using the product rule and the chain rule, we get:

x(dy/dy) + y + 7e^y(dy/dy) = 0

Simplifying, we get:

x + y + 7e^y = 0

Now, let's differentiate this equation with respect to y again to find the second derivative:

d²/dy²(x + y + 7e^y) = 0

Since we are looking for the second derivative at the point where x=0, we can ignore the x term.

Differentiating the remaining terms, we get:

d²y/dy² + e^y = 0

This is the second derivative at the point where x=0.

User Norris Oduro
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