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.

The derivative of the logarithm with base "e" is a fascinating subject in calculus. Look at this table, it might spark your curiosity:

We’ll show now that the spot with ? should be filled with the natural logarithm of x. You can either watch the video or read below.

Some tools that we’ll use are the properties of the logarithm:

We’ll also use the following property of continuous functions:

that is, that we can exchange the order between a continuous function and a limit sign. To find the derivative of the natural logarithm we’ll first write the definition of the derivative:

And now we apply the second property of logarithms in the list:

The argument of the logarithm we can also write as:

So, we have that:

Now we use the third property of logarithms:

Now we’ll use the clever substitution method we use to solve limits involving number e. We make:

This is equivalent to:

Let’s note something now. We are working with a limit where x approaches 0. If x remains fixed and Delta x approaches 0, the fraction Delta x/x will approach zero. That means that 1/n approaches zero (because it equals Delta x/x) That could only be possible if n approaches infinity.

All that we said resumes to:

So, doing the substitutions in our limit, we’ll have a function of the variable n:

We can also write this as:

Now we apply property three of logarithms again:

Because x is not a function of n we can take it out of the limit sign:

Now we’ll use the property of continuous functions that I noted above.

This property lets us change place between the function and the limit sign:

The argument of the natural logarithm is the definition of number e:

So, we’re left with:

But lne = 1. So, finally:

This result is pretty amazing, and quite unexpected. Let’s solve now some problems related to this concept.

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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