Skip to main content

5. Derivative of the Logarithmic Function

by M. Bourne

Later On this Page

First, let's look at a graph of the log function with base e, that is:

f(x) = loge(x) (usually written "ln x").

The tangent at x = 2 is included on the graph.

The slope of the tangent of y = ln x at x = 2 is 1/2. (We can observe this from the graph, by looking at the ratio rise/run).

If y = ln x,

x 1 2 3 4 5
slope of graph 1 1/2 1/3 1/4 1/5
1/x 1 1/2 1/3 1/4 1/5

We see that the slope of the graph for each value of x is equal to 1/x. This works for any positive value of x (we cannot have the logarithm of a negative number, of course).

If we did many more examples, we could conclude that the derivative of the logarithm function y = ln x is

dy/dx = 1/x

Note 1: Actually, this result comes from first principles.

Note 2: We are using logarithms with base e. If you need a reminder about log functions, check out Log base e from before.

Continues below

Derivative of the Logarithm Function y = ln x

The derivative of the logarithmic function y = ln x is given by:

d/(dx)(ln\ x)=1/x

You will see it written in a few other ways as well. The following are equivalent:

d/(dx)log_ex=1/x

If y = ln x, then (dy)/(dx)=1/x

We now show where the formula for the derivative of log_e x comes from, using first principles.

Proof of formula

For this proof, we'll need the following background mathematics.

First principles formula for the derivative of a function f(x), that is:

(df)/(dx) = lim_{h->0}(f(x+h)-f(x))/h

The logarithm laws

log a - log b= log (a/b)

and

n log a = log a^n.

The fact "log" and e are inverses, so

log_e(e) = 1.

The well-known limit

lim_{t->0}(1+t)^{1"/"t} = e

(We will not prove this limit here. The graph on the right demonstrates that as t->0, the graph of y=(1+t)^{1"/"t} approaches the value e~~2.71828.)

I will write log(x) to mean log_e(x) = ln(x), to make it easier to read.

We have f(x) = log(x) so the derivative will be given by:

(df)/(dx) = lim_{h->0}(log(x+h)-log(x))/h

Now the top of our fraction is

log(x+h)-log(x)  = log((x+h)/x)  = log(1 + h/x).

To simplify the algebra, we now substitute t=h/x and this gives us h = xt. Of course,

lim_{h->0}(h) = lim_{t->0}(xt) = 0.

So we have now:

(log(x+h)-log(x))/h  = 1/(xt)log(1 + t).

We write this as:

1/x[1/tlog(1 + t)].

Considering the expression in square brackets, we use the log law

n log a = log a^n

and write:

1/tlog(1 + t) = log(1+t)^(1"/"t}

Next, the limit of the expression (1+t)^(1"/"t} is:

lim_{t->0}(1 + t)^{1"/"t} = e.

Because log(e)=1, we can conclude:

(dlog(x))/(dx) = lim_{h->0}(log(x+h)-log(x))/h

= 1/x log(lim_{t->0}(1 + t)^{1"/"t})

 = 1/x log(e)

 = 1/x

Please support IntMath!

Tip

For some problems, we can use the logarithm laws to simplify our log expression before differentiating it.

Example 1

Find the derivative of

y = ln 2x

Answer

We use the log law:

log ab = log a + log b

We can write our question as:

y = ln 2x = ln 2 + ln x

Now, the derivative of a constant is 0, so

d/(dx)ln\ 2=0

So we are left with (from our formula above)

d/(dx)(ln\ x)=1/x

The final answer is:

(dy)/(dx)=1/x

We can see from the following graph that the slope of y = ln 2x (curve in green, tangent in magenta) is the same as the slope of y = ln x (curve in gray, tangent in dashed gray), at the point x = 2.

Example 2

Find the derivative of

y = ln x2

Answer

We use the log law:

log an = n log a

So we can write the question as

y = ln x2 = 2 ln x

The derivative will be simply 2 times the derivative of ln x.

So the answer is:

d/(dx)ln\ x^2=2 d/(dx)ln\ x=2/x

We can see from the graph of y = ln x2 (curve in black, tangent in red) that the slope is twice the slope of y = ln x (curve in blue, tangent in pink).

NOTE: The graph of y=ln(x^2) actually has 2 "arms", one on the negative side and one on the positive. The above graph only shows the positive arm for simplicity.

Please support IntMath!

Derivative of y = ln u (where u is a function of x)

Unfortunately, we can only use the logarithm laws to help us in a limited number of logarithm differentiation question types.

Most often, we need to find the derivative of a logarithm of some function of x. For example, we may need to find the derivative of y = 2 ln (3x2 − 1).

We need the following formula to solve such problems.

If

y = ln u

and u is some function of x, then:

(dy)/(dx)=(u')/u

where u' is the derivative of u

Another way to write this is

(dy)/(dx)=1/u(du)/(dx)

You might also see the following form. It means the same thing.

If

y = ln f(x),

then the derivative of y is given by:

(dy)/(dx)=(f'(x))/(f(x)

Example 3

Find the derivative of

y = 2 ln (3x2 − 1).

Answer

We put

u = 3x2 − 1

Then the derivative of u is given by:

u'=(du)/dx=6x

So the final answer is :

(dy)/(dx)=2 (u')/u

=2xx(6x)/(3x^2-1)

=(12x)/(3x^2-1)

Example 4

Find the derivative of

y = ln(1 − 2x)3.

Answer

First, we simplify our log expression using the log law:

log an = n log a

We can write

y = ln(1 2x)3 = 3 ln(1 2x)

Then we put

u = 1 − 2x

So

u' = (du)/dx=-2

So our answer is:

(dy)/(dx)=3 (u')/u

=3xx(-2)/(1-2x)

=(-6)/(1-2x)

Example 5

Find the derivative of y=ln[(sin 2x)(sqrt(x^2+1))]

Answer

First, we use the following log laws to simplify our logarithm expression:

log ab = log a + log b

and

log an = n log a

So we can write our question as:

y=ln[(sin 2x)(sqrt(x^2+1))]

=ln(sin 2x)+ln(sqrt(x^2+1))

=ln(sin 2x)+ln(x^2+1)^(1/2)

=ln(sin 2x)+1/2ln(x^2+1)

Next, we use the following rule (twice) to differentiate the two log terms:

(dy)/(dx)=(u')/u

For the first term,

u = sin 2x

u' = 2 cos 2x

For the second term, we put

u = x2 + 1,

giving

u' = 2x

So our final answer is:

(dy)/(dx)=(2\ cos 2x)/(sin 2x)+1/2 (2x)/(x^2+1)

=2\ cot\ 2x+x/(x^2+1)

Easy to understand math videos:
MathTutorDVD.com

Differentiating Logarithmic Functions with Bases other than e

If

u = f(x) is a function of x,

and

y = logb u is a logarithm with base b,

then we can obtain the derivative of the logarithm function with base b using:

(dy)/(dx)=(log_be)(u')/u

where

u' is the derivative of u

logbe is a constant. See change of base rule to see how to work out such constants on your calculator.)

Note 1: This formula is derived from first principles.

Note 2: If we choose e as the base, then the derivative of ln u, where u is a function of x, simply gives us our formula above:

(dy)/(dx)=(u')/u

[Recall that logee = 1. ]

[See the chapter on Exponential and Logarithmic Functions base e if you need a refresher on all this.]

Example 6

Find the derivative of y = log26x.

Answer

We begin by using the following log rule to simplify our question:

log ab = log a + log b

We can write our question as:

y = log26x = log26 + log2x

The first term, log26, is a constant, so its derivative is 0.

The derivative of the second term is as follows, using our formula:

(dy)/(dx)=(log_2e) (1/x)=(log_2e)/x

The term on the top, log2e, is a constant. If we need a decimal value, we can work it out using change of base as follows:

log_2e=(log_10e)/(log_10 2)=1.442695041

Example 7

Find the derivative of y = 3 log7(x2 + 1).

Answer

We put

u = x2 + 1,

giving

u' = 2x

Applying the formula, we have:

(dy)/(dx)=3\ (log_7e)(2x)/(x^2+1)

=6\ (log_7e) x/(x^2+1)

=3.083 x/(x^2+1)

The value 3.083 comes from using the change of base rule.

Note: Where possible, always use the properties of logarithms to simplify the process of obtaining the derivatives.

Exercises

1. Find the derivative of

y = ln(2x3x)2.

Answer

Using the Log Law log a^n = n log a, we can write:

y = ln(2x3x)2 = 2 ln(2x3x)

Put

u = 2x^3 − x

so

u' = 6x^2 − 1

This gives us:

(dy)/(dx)= (dy)/(du)(du)/(dx)  =2(6x^2-1)/(2x^3-x

x ≠ ±sqrt(0.5),

x ≠ 0

NOTE: We need to be careful with the domain of this solution, as it is only correct for certain values of x.

The graph of y = ln(2x3x)2 (which has power 2) is defined for all x except

 Âħsqrt(0.5), 0

Its graph is as follows:

The graph of y = 2 ln(2x3x), however, (it has 2 × at the front) is only defined for a more limited domain (since we cannot have the logarithm of a negative number.)

So we can only have x in the range -sqrt 0.5 < x < 0 and x > sqrt0.5.

So when we find the differentiation of a logarithm using the shortcut given above, we need to be careful that the domain of the function and the domain of the derivative are stated.

2. Find the derivative of

y = ln(cos x2).

Answer

Firstly,

d/(dx)cos x^2=-2x\ sin x^2

So

(dy)/(dx)=(-2x\ sin x^2)/(cos x^2)=-2x\ tan x^2

Get the Daily Math Tweet!
IntMath on Twitter

3. Find the derivative of

y = x ln3 x.

Answer

The notation

y = x(ln^3 x)

means

y = x(ln x)^3

Note that we cannot use the log law

log a^n= n log a

Our expression is not

y = x\ ln x^3

The brackets make all the difference!

This is a product of x and (ln x)^3. So

(dy)/(dx)=x(3(ln x)^2)/x+(ln x)^3(1)

=3(ln x)^2+(ln x)^3

=(ln x)^2(3+ln x)

Get the Daily Math Tweet!
IntMath on Twitter

4. Find the derivative of

3 ln xy + sin y = x2.

Answer

We need to recognise that this is an implicit function. We can simplify it first:

3(ln\ x+ln\ y)+sin y=x^2

Taking derivatives:

3(1/x+1/y(dy)/(dx))+cos y (dy)/(dx)=2x

Collecting terms gives us:

3/y(dy)/(dx)+cos y(dy)/(dx)=2x-3/x

(dy)/(dx)(3/y+cos y)=2x-3/x

So

(dy)/(dx)=(2x-3/x)/(3/y+cos y)

=(2x^2y-3y)/(3x+xy\ cos y)

Easy to understand math videos:
MathTutorDVD.com

5. Find the derivative of

y = (sin x)x

by first taking logarithms of each side of the equation.

Answer

NOTE: This has an exponent which is variable. We cannot use our formula

d/dx x^n=nx^(n-1)

from before.

Now

ln\ y=ln[(sin x)^x]=x\ ln(sin x)

So

1/y(dy)/(dx)=x(cos x)/(sin x)+ln(sin x)(1)

Multiplying through by y gives:

(dy)/(dx)=y(x\ cot\ x+ln(sin x))

=(sin x)^x(x\ cot\ x+ln(sin x))

The graph of the function in Exercise 5 is quite interesting:

Search IntMath

Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!

See the Interactive Mathematics spam guarantee.