Let y be defined implicitly by the equation ln(8y)=5xy.

Question

A) Use implicit differentiation to find the second derivative of y with respect to x. ((d^2)y)/(d(x^2)= ?

simplify your answer as much as possible.

B) Find the point on the curve where ((d^2)y)/(d(x^2)=0.

Answer 1

ln(8y) = 5xy [apply the derivative operator:

d/dx(ln(8y)) = d/dx(5xy)

Using the chain rule, the left side is 1/(8y)•8•dy/dx

Using the product rule the right side is 5y + 5x•dy/dx

1/(8y)•8•dy/dx = 5y + 5x•dy/dx

(1/y)•dy/dx - 5x•dy/dx = 5y

dy/dx(1/y - 5x) = 5y

dy/dx(1/y - 5xy/y) = 5y

dy/dx(1 - 5xy)/y = 5y

dy/dx = 5y•(y/(1 - 5xy)) = 5y²/(1 - 5xy)

To find the second derivative, take the derivative of the first derivative:

Apply the derivative operator:

d²y/dx² = d/dx(5y²/(1 - 5xy))

Use the quotient rule

(u'v - uv')/v² where u = 5y² and v = 1 - 5xy

u' = 10y•dy/dx = 10y•(5y²/(1 - 5xy)) = 50y³/(1 - 5xy)

v' (from the product rule):

v' = -5y + -5x•dy/dx

v' = -5y + -5x•5y²/(1 - 5xy)

v' = -5y - 25xy²/(1 - 5xy)

v' = -5y(1 - 5xy)/(1 - 5xy) - 25xy²/(1 - 5xy)

v' = (-5y + 25xy²)/(1 - 5xy) - 25xy²/(1 - 5xy)

v' = -5y/(1 - 5xy)

So, (u'v - uv')/v² = [(50y³/(1 - 5xy))(1 - 5xy) - (5y²)(-5y/(1 - 5xy))]/(1 - 5xy)²

d²y/dx² = [(50y³ - 250xy⁴)/(1 - 5xy) + 25y³/(1 - 5xy)]/(1 - 5xy)²

d²y/dx² = [(50y³ - 250xy⁴ + 25y³)/(1 - 5xy)]/(1 - 5xy)²

d²y/dx² = (75y³ - 250xy⁴)/(1 - 5xy)³

d²y/dx² = 25y³(3 - 10xy)/(1 - 5xy)³

To find the point on the curve where d²y/dx² = 0, make 25y³(3 - 10xy)/(1 - 5xy)³ = 0

25y³(3 - 10xy)/(1 - 5xy)³ = 0 only when the numerator = 0

25y³(3 - 10xy) = 0 when either 25y³ = 0 (or y = 0) or when 3 - 10xy = 0

3 - 10xy = 0

3 = 10xy

x = 0.3/y

Going back to the original equation: ln(8y) = 5xy and looking at the solution y = 0, we see that y cannot be 0 so we can throw out that solution.

For x = 0.3/y we can plug that in:

ln(8y) = 5(0.3/y)y

ln(8y) = 5(0.3) = 1.5

e^1.5 = 8y

y = (e^1.5)/8 and since x = 0.3/y then x = 0.3/(e^1.5)/8 = 2.4/e^1.5 so the point is (2.4/e^1.5, e^1.5/8) [which is approximately (0.536, 0.560)]