Do you know the formula for the derivative of natural log? And do you know why the natural logarithm is "natural"? In this page we'll see the intuition behind this formula and learn how to solve derivatives using it.
Do you have a question or doubt about this topic? An "impossible problem"? Submit it here!
The derivative of the logarithm with base "e" is a fascinating subject in calculus. Look at this table, it might spark your curiosity:

Ok, let's get to work. You can first watch this video or read below:
We'll simply use the definition of the derivative:

Here we can apply the following property of logarithms:

Applied to our derivative:


We can also write this as:

Now we'll use this property of logs:

That is:

Now we'll use a clever substitution:

This is equivalent to:

You'll see in a moment why we use this substitution. Now, we'll note something of value. We hava a limit with Δx approaching 0. What does n approach when Δx approaches 0?

As Δx approaches 0, 1 over n approaches 0. That's equivalent to saying that n approaches infinity.
So, doing these substitutions, our derivative becomes:

We can also write this as:

Using the properties of logs again:

Now, taking the 1 over x out of the limit sign, because it doesn't depend on n:

Because the log function is continuous, we can write this as:

(If this confuses you, please review the page on continuous functions).
The limit that's inside the log function is the definition of number e:

So, the derivative simplifies to:

And the natural log of e is just one. So we're left with:

Pretty amazing result.
Let's solve some derivatives involving the natural logarithm. Let's find the derivative of:

Looks complicated? Here we'll apply the chain rule. What is the derivative of the outside function? It is one over the argument:

And the derivative of cos(x) is -sin(x):

Let's derive:

We'll apply the chain rule:

And the derivative of the inside function is just a:

Let's derive:

What would happen if you applied the chain rule directly? This would become a very long and hairy problem.
Instead, we'll use a trigonometric trick. Inside the square root, let's multiply both the numerator and denominator by 1+sin(x):

What we have in the denominator is cos squared of x. So:

This looks much easier, doesn't it? Now we can use the chain rule:


Now we'll apply the product rule in the second factor:

And now it is just algebra:

Performing the sum:

We can simplify a bunch of things:

And finally we have:


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