149k views
0 votes
If xy + 3eʸ = 3e, find the value of y'' at the point where x = 0. A thorough explanation of how to do it would be really great. I know it is implicit differentiation, just don't know how to apply it.

User Samball
by
8.7k points

1 Answer

4 votes

To find the value of y'' at the point where x = 0, we'll need to differentiate the given equation twice with respect to x and then evaluate it at x = 0. Let's start by finding the first and second derivatives of y with respect to x.

Given equation: xy + 3e^y = 3e

1. Find the first derivative (y'):

Differentiate both sides of the equation with respect to x:

d/dx [xy + 3e^y] = d/dx [3e]

Using the product rule for the left side (d(uv)/dx = u'v + uv'), where u = x and v = y:

x * (dy/dx) + y * (dx/dx) + 3e^y * (dy/dx) = 0

Simplifying:

x(dy/dx) + y + 3e^y(dy/dx) = 0

Now, isolate dy/dx:

(dy/dx)(x + 3e^y) = -y

dy/dx = -y / (x + 3e^y)

2. Find the second derivative (y''):

Now, differentiate dy/dx with respect to x again:

d/dx [-y / (x + 3e^y)] = ?

To find this derivative, we can use the quotient rule (d(u/v)/dx = (v * du/dx - u * dv/dx) / v^2):

= [(x + 3e^y)(d/dx(-y)) - (-y)(d/dx(x + 3e^y))] / (x + 3e^y)^2

Now, we can substitute the expression for dy/dx and simplify:

= [(x + 3e^y)(-dy/dx) - (-y)(1 + 3e^y(dy/dx))] / (x + 3e^y)^2

= [(x + 3e^y)(y / (x + 3e^y)) + y(1 + 3e^y(-y / (x + 3e^y))))] / (x + 3e^y)^2

= [xy + 3e^y + y - 3e^(2y)] / (x + 3e^y)^2

Now, we need to evaluate y'' at the point where x = 0:

y''(0) = [0 + 3e^y + y - 3e^(2y)] / (0 + 3e^y)^2

= (3e^y + y - 3e^(2y)) / (3e^y)^2

= (3e^y + y - 3e^(2y)) / (9e^(2y))

This is the expression for y'' at the point where x = 0 in terms of y. Further simplification or numerical evaluation would require knowledge of the specific value of y at that point.

User Arbuzov
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories