how do you find the derivative when there is ln? What is the derivative of ln(natural log)?

Question

Like in this ex. y= ln^(5√x) does ln mean 1/ something?

Answer 1

The ln(a) where a is an arbitrary number returns the number x for which e^x = a. This of course assumes that a is in the domain of the natural log function, in other words a > 0. This comes from the fact that there exists no real number x for which e^x is either 0 or less than zero.

The derivative of ln(u) where u is a differentiable function of x is u'/u

Answer 2

What helps me to remember is that when you run into a problem like this where you have to take the derivative, think: Derivative over what you're taking the derivative of.

Such as:

ln(5x+3)

The derivative is:

5/(5x+3).

Its the same thing as multiplying the derivative by 1 over what you're taking the derivative of.

ln(5sqrtx)

5(x^(1/2))

5/2x^-1/2/(5x^(1/2))

Answer 3

Not sure that you have typed question correctly.
Consider the following example:-
y = ln ( 5√x )
let u = 5√x = 5 x^(1/2)
du/dx = (5/2).x^(- 1/2) = (5 / 2√x)
y = ln u
dy/du = 1 / u = 1 / (5√x)
dy/dx = (dy/du).(du/dx)
dy/dx = (1 / 5√x).(5 / 2√x)
dy / dx = 1 / (2x)

Answer 4

if y = ln(u)
y' = u'/u

y = ln^(5√x)
y = ln[√(25x)]
y = ln[(25x)^(1/2)]
y = (1/2)ln(25x)
y' = (1/2)(1/(25x))(25)
y' = (1/2)(1/x)

y = ln^(5√x)
y' = (1/(5√x))(5)(1/2)(1/√x)
y' = (1/2)(1/x)

Answer 5

Chain Rule: y = ln(csc(5x)) y' = [1/csc(5x)]*[-csc(5x)cot(5x)]*[5] = -5cot(5x)