# Derivative of ln x

How to find the derivative of the given function by using the derivative of ln x: formulas, 5 examples, and their solutions.

## Formulas

### Derivative of ln x

The derivative of ln x, [ln x]',
is 1/x.

Natural Logarithm

### Derivative of ln |x|

The derivative of ln |x|, [ln |x|]',
is also 1/x.

## Example 1

### Solution

y = ln (2x + 7) is a composite function of
y = ln (whole) and (whole) = 2x + 7.

So the derivative of the composite function is,
the derivative of the outer function ln (2x + 7), 1/(2x + 7)
times,
the derivative of the inner function 2x + 7, 2.

Arrange the expression.

So 2/(2x + 7) is the derivative of ln (2x + 7).

## Example 2

### Solution

y = ln |x3 - 2x| is a composite function of
y = ln |(whole)| and (whole) = x3 - 2x.

So the derivative of the composite function is,
the derivative of the outer function ln |(x3 - 2x)|, 1/(x3 - 2x)
times,
the derivative of the inner function x3 - 2x, 3x2 - 2.

Arrange the expression.

So (3x2 - 2)/(x3 - 2x) is the derivative of ln |x3 - 2x|.

## Example 3

### Solution

Take the exponent of the x4
out from the ln.

Logarithm of a Power

When taking out the exponent,
write the absolute value to the x,
because the number in the log cannot be minus.

y = 4 ln |x|

Then y' = 4⋅(1/x).

Arrange the expression.

So 4/x is the derivative of ln x4.

## Example 4

### Solution

x3 ln x is the product of x3 and ln x.

So the derivative of the product is,
the derivative of x3, 3x2
times ln x
plus
x3
times, the derivative of ln x, 1/x.

+x2⋅(1/x) = +x2

(3x2)(ln x) + x2 = x2(3 ln x + 1)

Common Monomial Factor

So x2(3 ln x + 1) is the derivative of x3 ln x.

## Example 5

### Solution

ln both sides.
Then ln |y| = ln |xx|.

The number in the log cannot be minus.
So write the absolute value signs.

Take the exponent x out from the absolute value sign.

Take the exponent x out from the ln.
Then ln |y| = x ln |x|.

Differentiate both sides.

The left side ln |y| becomes (1/y)⋅y'.

Derivative of an Implicit Function

The right side is the product of x and ln |x|.

So the derivative of x ln |x| is,
the derivative of x, 1
times ln |x|
plus
x
times, the derivative of ln |x|, 1/x.

So ln |y| = x ln |x| becomes
(1/y)⋅y' = 1⋅ln |x| + x⋅(1/x).

(1/y)⋅y' = y'/y

1⋅ln |x| = ln |x|
+x⋅(1/x) = +1

Multiply y to both sides.
Then y' = y(ln |x| + 1).

The given function is y = xx.
Put this into y' = y(ln |x| + 1).

Then y' = xx(ln |x| + 1).

So xx(ln |x| + 1) is the derivative of xx.