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

Example

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

Example

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

Example

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

Example

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

Example

Solution

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.