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.

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`

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.

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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.

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)`

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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)`

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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)`

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`

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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 − 1`

This gives us:

`(dy)/(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.

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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`

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)`

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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)`

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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: