Implicit Derivatives

Question

The equation given is ln(8y)=5xy .

Find dy/dx Then d^2y/dx^2. Also find the point on the curve where d^2y/dx^2=0 point (x,y).

Answer 1

Below I use y' for dy/dx and y'' for d²y/dx² in equations used to find implicit derivatives.

Finding the first derivative dy/dx

If we take ln(8y) = 5xy and solve for x we get

x = ln(8y)/5y which is a much easier equation to work with

Taking the derivative of both sides (dx/dx = 1), we get

1 = [5y(y’/y) – 5y’ ln(8y)]/(25y²) = [y’(1 – ln(8y)]/(5y²) which means that

y’ = 5y²/(1 – ln(8y))

Finding the second derivative d²y/dx²

y'' = [(1 – ln(8y))10yy’ – 5y²(–(y’/y)]/(1 – ln(8y))² = (10yy’(1 – ln(8y)) + 5yy’)/(1 – ln(8y))²

Using the common factor 5yy’, we can factor and simplify the numerator

(10yy’(1 – ln(8y)) + 5yy’) = 5yy’(2(1 – ln(8y)) +1) = 5yy’(3 – 2ln(8y))

So

y'' = 5yy’(3 – 2ln(8y))/(1 – ln(8y))²

Substituting y’ = 5y²/(1 – ln(8y)) gives

y'' = 5y[5y²/(1 – ln(8y))](3 – 2ln(8y))/(1 – ln(8y))² = [25y³(3 – 2ln(8y))]/(1 – ln(8y))³

Finding the point (x,y) where d²y/dx² = 0

y'' = 0 when the numerator = 0, and

25y³(3 – 2ln(8y)) = 0 when y = 0 or 3 – 2ln(8y) = 0

We can throw out y = 0 because ln(8y) is undefined unless y > 0

So the y value of the point (x,y) where y'' = 0 comes from 3 – 2ln(8y) = 0

This means

2ln(8y) = 3, so

ln(8y) = 3/2

Taking the exponential of each side gives

8y = e^3/2 which means that y = e^3/2/8

We get the x value by taking x = ln(8y)/5y and substituting this value of y

x = ln(8×e^3/2/8)/(5×e^3/2/8) = ln(e^3/2)/(5/8×e^3/2) = (3/2)/[5/8×e^3/2)

So x = (3/2)(8/5)e^–3/2 = (12/5)e^–3/2 = 2.4e^–3/2

Answers

dy/dx = 5y²/(1 – ln(8y))

d²y/dx² = [25y³(3 – 2ln(8y))]/(1 – ln(8y))³



d²y/dx² = 0 at the point (2.4e^–3/2, e^3/2/8)