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AP Calculus AB & BC: Help and Review17 chapters | 160 lessons

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Lesson Transcript

Instructor:
*Laura Pennington*

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

The derivative of ln(x) is a well-known derivative. This lesson will show us the steps involved in finding this derivative, and it will go over a real world application that involves the derivative of ln(x).

We want to find the derivative of ln(*x*). The derivative of ln(*x*) is 1/*x*, and is actually a well-known derivative that most put to memory. However, it's always useful to know where this formula comes from, so let's take a look at the steps to actually find this derivative.

To find the derivative of ln(*x*), the first thing we do is let *y* = ln(*x*). Next, we use the definition of a logarithm to write *y* = ln(*x*) in logarithmic form. The definition of logarithms states that *y* = log *b* (*x*) is equivalent to *b* *y* = *x*. Therefore, by the definition of logarithms and the fact that ln(*x*) is a logarithm with base *e*, we have that *y* = ln(*x*) is equivalent to *e* *y* = *x*.

Okay, just a few more steps, and we'll have our formula! The next thing we want to do is treat *y* as a function of *x*, and take the derivative of each side of the equation with respect to *x*. We use the chain rule on the left hand side of the equation to find the derivative. The **chain rule** is a rule we use to take the derivative of a composition of functions, and it has two forms.

The left hand side of the equation is *e* *y*, where *y* is a function of *x*, so if we let *f*(*x*) = *e* *x* and *g*(*x*) = *y*, then *f*(*g*(*x*)) = *e* *y*. Since the derivative of *e* to a variable (such as *e* *x*) is the same as the original, the derivative of *f'(g(x))* is *e* *y*. Therefore, by the chain rule, the derivative of *e* *y* is *e* *y* *dy*/*dx*. On the right hand side we have the derivative of *x*, which is 1.

We have (*e* *y*) *dy*/*dx* = 1. Now, recall that *e* *y* = *x*. We're going to use this fact to plug *x* into our equation for *e* *y*.

This gives us the equation (*x*)*dy*/*dx* = 1. We're getting super close now! Are you as excited as I am? We can divide both sides of this equation by *x* to get *dy*/*dx* = 1/*x*. The last thing is to recall that *y* = ln(*x*) and plug this into our equation for *y*.

Ta-da! Now, we see that *d*/*dx* ln(*x*) = 1/*x*, and now we know why this formula for the derivative of ln(*x*) is true. So what's our solution? The derivative of ln(*x*) is 1 / *x*.

As we said, the derivative of ln(*x*) is a well-known derivative that most put to memory. This is because this derivative shows up often in real world applications. Therefore, it's very useful to know the derivative so that you don't have to go through the process of finding it every time it comes up.

For example, consider a certain plane that takes off at sea level. The plane's altitude (in feet) at *x* minutes can be given by the function *h*(*x*) = 2000ln(*x*).

The derivative of a function represents a rate of change. Therefore, the derivative of *h* represents the rate of the plane's climb at *x* minutes. Let's find that derivative so we'll have a formula for the plane's climb rate.

Because *h* ' (*x*) = 2000ln(*x*), we see that *h* ' (*x*) = 2000 / *x*. This function represents the plane's climb rate in feet per minute at *x* minutes. For example, if we want to know the climb rate two minutes after the plane takes off, we plug 2 into the derivative to get the following.

We get that *h* ' (2) = 1000. This tells us that the plane is climbing at a rate of 1000 feet per minute two minutes after the plane has taken off.

Pretty awesome, isn't it? We know how to find the derivative of ln(*x*), and we also see why it is useful to put the derivative to memory: because it makes solving problems where it comes up much easier. Thus, if you take one thing from this lesson, let it be that the derivative of ln(*x*) is 1 / *x*!

Let's take a few moments to briefly recap what we've learned about solving the derivative of ln(*x*). The steps are as follows:

- Let
*y*= ln(*x*). - Use the definition of a logarithm to write
*y*= ln(*x*) in logarithmic form. This definition states that*y*= log*b*(*x*) is equivalent to*b**y*=*x*, so therefore,*y*= ln(*x*) is equivalent to*e**y*=*x*. - Treat
*y*as a function of*x*, and take the derivative of each side of the equation with respect to*x*. - Use the chain rule on the left hand side of the equation to find the derivative. Recall that the
**chain rule**is a rule we use to take the derivative of a composition of functions, and it has two forms. This will lead us to the conclusion that*e**y*is*e**y**dy*/*dx*. The derivative of*x*is 1. - Plug
*x*into our equation for 'e*%.* - We get (
*x*)*dy/dx*= 1 and divide both sides of this equation by*x*to get*dy/dx*= 1/*x*. - Plug
*y*= ln(*x*) for*y*. - See that
*d/dx*ln(*x*) = 1/*x*and, therefore, that the derivative of ln(*x*) is 1/*x*.

There, simple, right?

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AP Calculus AB & BC: Help and Review17 chapters | 160 lessons

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