# Derivative of ln x

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

## FormulaDerivative of ln x

The derivative of ln x, [ln x]',

is 1/x.

Natural Logarithm

## FormulaDerivative of ln |x|

The derivative of ln |x|, [ln |x|]',

is also 1/x.

## Exampley = ln (2x + 7), y' = ?

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).

## Exampley = ln |x^{3} - 2x|, y' = ?

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

is a composite function of

y = ln |(whole)| and (whole) = x^{3} - 2x.

So the derivative of the composite function is,

the derivative of the outer function ln |(x^{3} - 2x)|, 1/(x^{3} - 2x)

times,

the derivative of the inner function x^{3} - 2x, 3x^{2} - 2.

Arrange the expression.

So

(3x^{2} - 2)/(x^{3} - 2x)

is the derivative of ln |x^{3} - 2x|.

## Exampley = ln x^{4}, y' = ?

Take the exponent of the x^{4}

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 x^{4}.

## Exampley = x^{3} ln x, y' = ?

x^{3} ln x

is the product of x^{3} and ln x.

So the derivative of the product is,

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

times ln x

plus

x^{3}

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

+x^{2}⋅(1/x) = +x^{2}

(3x^{2})(ln x) + x^{2} = x^{2}(3 ln x + 1)

Common Monomial Factor

So

x^{2}(3 ln x + 1)

is the derivative of x^{3} ln x.

## Exampley = x^{x}, y' = ?

ln both sides.

Then ln |y| = ln |x^{x}|.

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 = x^{x}.

Put this into y' = y(ln |x| + 1).

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

So

x^{x}(ln |x| + 1)

is the derivative of x^{x}.