# Derivative of ln *x*

How to solve the derivative of ln x problems: formulas (ln x, ln |x|), proofs, examples, and their solutions.

## Formula: [ln *x*]'

The derivative of ln *x* is 1/*x*.

## Formula: [ln |*x*|]'

The derivative of ln |*x*| is also 1/*x*.

## Proofs

Let's prove [ln *x*]' = 1/*x*.

Use the definition of a derivative function.

Move the denominator *h* to the front part.

Split the base into two fractions

and simplify *x*/*x*.

Move 1/*h* into the base's exponent.

Logarithms of powers

The base is 1 + *h*/*x*.

So, change the exponent 1/*h* into *x*/*h*.

(the reciprocal of *h*/*x*)

And, to undo the *x*,

multiply 1/*x* to the exponent.

Take the exponent 1/*x* out.

Logarithms of powers

As *h* → 0,

the gray part goes to *e*.

Definition of *e* (Mathematical constant)

So (right side) = (1/*x*) ln *e*.

So [ln *x*]' = 1/*x*.

Case 1: *x* > 0

|*x*| = *x*

So [ln |*x*|]' = [ln *x*]'.

= 1/*x*.

So [ln |*x*|]' = 1/*x*

when *x* > 0.

Case 2: *x* < 0

|*x*| = -*x*

So [ln |*x*|]' = [ln (-*x*)]'.

And [ln (-*x*)]' = [1/(-*x*)]⋅(-1).

(Multiplied -1 is the derivative of -*x*.)

Chain rule in differentiation

So [ln |*x*|]' = 1/*x*

when *x* < 0.

So [ln |*x*|]' = 1/*x*.

## Example 1

*y* = ln (2*x* + 7)

So *y*' is equal to,

the derivative of the outer part, 1/(2*x* + 7)

times, the derivative of the inner part, 2.

Chain rule in differentiation

Derivative of polynomials

## Example 2

*y* = ln |*x*^{3} - 2*x*|

So *y*' is equal to,

the derivative of the outer part, 1/(*x*^{3} - 2*x*)

times, the derivative of the inner part, (3*x*^{2} - 2).

Chain rule in differentiation

Derivative of polynomials

## Example 3

Take the exponent 4 out.

Logarithms of powers

*y* = 4 ln |*x*|

So *y*' = 4⋅(1/*x*).

## Example 3: Different Solution

Let's solve this example differently.

Think *y* = ln *x*^{4} as a composite function:*y* = ln (*x*^{4}).

Then *y*' is equal to,

the derivative of the outer part, 1/*x*^{4}

times, the derivative of the inner part, 4*x*^{3}.

Chain rule in differentiation

Power rule in differentiation (Part 1)

(1/*x*^{4})⋅4*x*^{3} = 4/*x*

As you can see,

you can get the same answer: 4/*x*.

## Example 4

*y* = (*x*^{3})(ln *x*)

So *y*' is equal to,

the derivative of *x*^{3}, 3*x*^{2}

times ln *x*

plus *x*^{3},

times, the derivative of ln *x*, 1/*x*.

Product rule in differentiation

Derivative of polynomials

## Example 5

ln both sides,

covering the bases with the absolute value signs.

(The absolute value signs are added

becasue both sides can be (-).)

Natural logarithms

Differentiate both sides.

Implicit differentiation

Product rule in differentiation

Multiply *y* on both sides.

Change *y* into *x*^{x}.

Then *y*' = *x*^{x}(ln |*x*| + 1).