Question

What are the first and second derivatives of #f(x)=ln((4x-5)^(1/3)/(3x+8)^2) #?

Answer 1

`dy/dx = 4/(3(4x-5)) - 6/(3x + 8)`
`(d^2y)/dx^2 = -8/(3(4x-5)^2) + 9/(3x+8)^2`

Explanation:

Use the log properties to simplify

`y = ln((4x - 5)^(1/3)/(3x+8)^2)`

`y = ln((4x-5)^(1/3))-ln((3x+8)^2)`

`y = ln(4x-5)/3 - 2ln(3x+8)`

Say that `4x - 5 = u` and that `3x + 8 = v`

`y = ln(u)/3 - 2ln(v)`

`dy/dx = 1/3d/(du)ln(u)(du)/dx - 2d/(dv)ln(v)(dv)/dx`

`dy/dx = 1/3*1/ud/dx(4x-5) - 2*1/v*d/dx(3x+8)`

`dy/dx = 1/3*1/u*4 - 2*1/v*3`

`dy/dx = 4/(3(4x-5)) - 6/(3x + 8)`

To find the second derivative, just derivate again, we have

`dy/dx = 4/(3u) - 6/v`

`(d^2y)/dx^2 = 4/3*d/(du)1/u(du)/dx - 6d/(dv)1/v(dv)/dx`

`(d^2y)/dx^2 = 4/3*(-1/(2u^2))*4 - 6\*(-1/(2v^2))*3`

`(d^2y)/dx^2 = -8/(3(4x-5)^2) + 9/(3x+8)^2`