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# 5. Derivative of the Logarithmic Function

by M. Bourne

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.

## 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

### 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

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

(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

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.

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.

## 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).

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.

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

(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))]

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

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

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

## 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.

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).

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.

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

 ±sqrt(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).

Firstly,

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

So

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

3. Find the derivative of

y = x ln3 x.

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)

4. Find the derivative of

3 ln xy + sin y = x2.

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)

5. Find the derivative of

y = (sin x)x

by first taking logarithms of each side of the equation.

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:

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