5. Derivative of the Logarithmic Function

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:

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.

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.

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)`

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)`

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)`

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`

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.

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

y = ln(2x³ − x)² = 2 ln(2x³ − x)

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(2x³ − x)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(2x³ − x), 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.

Firstly,

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

So

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

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)`

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)`

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: