If xy + 8ey = 8e, find the value of y'' at the point where x = 0.

Question

If xy + 8e^y = 8e,

 find the value of y''

 at the point where x = 0.

Answer 1

It is not possible to solve the initial relation for y explicitly, so implicit differentiation must be used.

a) xy' + y + 8e^y(y') = 0

b) y'(x + 8e^y) = -y

c) y' = (-y)/(x + 8e^y)

At this point you can use the quotient rule on line c) to determine y'' or go back to line a) and determine y'' from there.

Let's use quotient rule on line c).

y'' = [(x+8e^y)(-1)y' - (-y)(1 + 8e^yy')]/(x+8e^y)²

At this point you could substitute the expression for y' into the expression for y'' and simplify. But, since the problem is to evaluate when x = 0, there is no requirement to get y'' in terms of x and y.

Note that when x = 0, from the original relation we have: 0 + 8e^y = 8e => e^y = e => y=1

So, (0,1) is the point where we want to evaluate y''.

Evaluating y'(0,1) results in: -1/(0+8e¹) = -1/(8e).

y''(0,1) = [(0+8e)(-1)(-1/8e) + (1)(1+8e(-1/8e))] / (8e)²

= [1 + 1 - 1] / 64e²

= 1/(64e²)

Since y''(0) is positive, that means the graph of the original relation is concave upward at the point (0,1).

This graph has a row showing y'' in terms of x and y.

desmos.com/calculator/ykjhbzxuqy