One of the questions that I sometimes made to myself when first studying calculus was: why the natural logarithm is "natural"? One of the reason it is called natural is the simplicity of the formula for the derivative of ln(x).

The formula for the derivative of ln(x) is not at all obvious. We know, because of the power rule, that the derivative of a polynomial is a polynomial. We also know we can obtain negative powers of x as derivatives of other negative powers of x.

To see what I mean, look at the following equations. In these equations, I just applied the power rule.

We can see from these equations that we can obtain any positive power of x as a derivative of a simple polynomial function. We can also obtain almost any negative power of x in this way. Now, what is the function whose derivaitve is xWe'll use a graphical method for the deduction of the derivatie of ln(x). For that, we'll use the geometric definition of derivative: the slope of the tangent line.

We'll begin with the graph of eThen, the graph of e

Now, when y approaches zero, we have that e

With respect to what line is the graph reflected? If you look closely at the graphs, you might be able to realize that they are reflected with respect to the diagonal line on the x-y plane. That is, the line y=x.

This means that to obtain the graph of ln(y), we just need to draw the graph of e

Now, let's return to what were trying to do. We want to calculate the derivative of ln(y). To do that, we need the slopes of the tangent lines. Now, as the graph of ln(y) is the reflection of the exponential function, its tangent lines will also be reflections.

This means that the tangent line to ln(y) at any point is the reflection of the corresponding tangent line to e

That is,

So, the corresponding tangent to ln(y) will have slope 1/e

Or if we return to the use of the more common letter x, we have that

That was the quick and easy graphical deduction of the derivative of ln(x). Now we are ready to apply it to calculate more derivatives.

Let's consider the function

Here we'll need to 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 calculate the derivative of

Here a and b are constants. We'll apply the chain rule:

And the derivative of the inside function is a:

Let's calculate the derivative of

We can actually solve this problem using the chain rule. However, if we go that route, this will become a long and hairy calculation.

Instead, we'll use an old trigonometric trick that will make our life easier. Inside the square root, let's multiply both the numerator and denominator by 1+sin(x):

The expression in the denominator is interesting. It actually equals cosThat simplified the expression. Now we can simplify it a little more, applying a property of logarithms

Now we are finally ready to calculate the derivative. We apply the chain rule

We can leave at that and accept that expression as the answer. In this case, though, we can simplify it

So the derivative equals

That's all for now. At this point you already know most rules for the calculation of derivatives. There is a nice trick, called implicit differentiation, that allows you to calculate the derivative of functions whose expression you don't know. And that's the next step from here: Implicit Differentiation.

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