In this lesson, we’ll review common derivatives and their rules, including the product, quotient and chain rules. We’ll also examine how to solve derivative problems through several examples.

## Review of Common Derivatives

Let’s take a few minutes to review the most commonly used derivatives. First, we have *x*^*n*th power.

The derivative of *x*^*n*th power is *n* times *x*^(*n*-1). Two similar derivatives are the derivatives (*d/dx*) of *x*, which is just 1, and *d/dx* of any constant, which is just 0. The derivative of the exponential *e*^*x* is *e*^*x*. It’s one of the most fantastic derivatives of all time because no matter how many times you differentiate, you get the same thing back: The derivative of *e*^*x* is *e*^*x*. The derivative of the natural log of *x* is 1/*x*.

Another class of commonly used derivatives are the trig functions. The derivative of sin(*x*) is cos(*x*), the derivative of cos(*x*) is -sin(*x*) and the derivative of the tangent of *x* is the secant squared of *x* (sec^2(*x*)).

## Review of the Rules of Derivatives

While you should memorize these six derivatives especially, always remember the most important lesson about derivatives: The derivative is the slope of the tangent to your function! Once you get those basics, there are three rules you should remember:First is the **product rule**.

Here let’s say that *f(x)* is the product of two other functions that depend on *x*. So here we’ve got *u* and *v*, which are both functions of *x*. The derivative of *f(x)* with respect to *x*, or *f`(x)*, is *uv`* + *vu`*, or the first times the derivative of the second, plus the second times the derivative of the first, where our first function is *u* and our second function is *v*.The second rule to remember is the **quotient rule**. So again, let’s use *u* and *v*.

If *f(x)*=*u*/*v*, then *f`(x)*=(*vu`* – *uv`*) / *v*^2. We can remember this by using our little rhyme: *Low d hi minus hi d low, all over the square of what’s below!*

The last rule to remember is the **chain rule**. You’re going to use the chain rule anytime you see parentheses or when you have composite functions, like *f(x)*=*g(h(x))*. In this case, *f`(x)*=*g`(h(x)) * h`(x).*

## First Example of Solving Derivatives

Let’s use these rules in an example. Let’s say *f(x)*=sin(*x*) / *x*^2.

Well since I have one function divided by another function, I’m going to use the quotient rule, which says *Low d hi minus hi d low, all over the square of what’s below!* Here, my low function is *x*^2, and my high function is sin(*x*). So let’s use this. *f'(x)* equals low (*x*^2), *d/dx* of high (sin(*x*)), minus high (sin(*x*)), *d/dx* low (*x*^2), all divided by what’s below, squared. So (*x*^2)^2. Well I know how to find *d/dx* of sin(*x*), that’s just one of the rules we know; the derivative of sin(*x*) is cos(*x*). For the derivative of *x*^2, I’m going to use my power rules.

*d/dx* of *x*^*n* is *nx*^(*n* – 1). So the derivative of *x*^2 is 2*x*^(2 – 1). Well, 2 – 1 is 1, so this is 2*x*^1, or just 2*x*.

So I can plug that in, and *f`(x)*=(*x^2*(cos(*x*)) – sin(*x*)(2*x*)) / (*x*^2)(*x*^2).Let’s simplify. Let’s move this 2*x* to the left side of sin(*x*), and let’s multiply *x*^2 * *x*^2 so that I get *x*^4. Well from here, you could leave it, but there’s an *x* in every single term.

So let’s factor out one *x* from the top so that I get *x*(*x*cos(*x*) – 2sin(*x*)). Let’s cancel this *x* with one of these four *x*s that we have on the bottom. With all of that, I find that the derivative of *f(x)* is (*x*cos(*x*) – 2sin(*x*)) / *x*^3.

## Second Example

Let’s do another example. Let’s say this time that *f(x)*=*e*^*x*^2. Because I have an exponential here, I kind of have implied parentheses.

So this is like saying *e*^(*x*^2). Since I see parentheses, I’m going to think chain rule. The chain rule says that I’m first going to find the derivative of this outside function, *e* to the something.

In this case, the something is *x*^2. Then I’m going to find the derivative of the inside, which is this *x*^2. So if I write this as *g(h(x))* is *e*^(*x*^2), then my *h(x)*, my inside function, is *x*^2. The derivative of *g* with respect to my *h(x)* here is *e*^(*x*^2). The derivative of *h(x)* is 2*x*.

Let’s plug that in. Since I’m using the chain rule, my derivative of *e*^(*x*^2) is *e*^(*x*^2) * the derivative of *x*^2, which is just 2*x*. So *f`(x)*=2*xe*^(*x*^2)

## Third Example

Let’s do another chain rule example. Let’s say we have the function *f(x)*=(cos(*x*))^3. We’ll usually write this as cos^3(*x*).

To find the derivative, *f`(x)*, we’re going to use the chain rule where our outside function is something cubed. In this case, that something is cos(*x*). Our outer function is that cubed. So the derivative (we’re going to use our power rule here, which says that if *f(x)*=*x*^*n*, then *f`(x)*=*nx*^(*n*-1).

Well here, *n* is 3, so our derivative is 3 times our *x* here, which is just parentheses, to the 3 – 1, or 2. What’s inside of the parentheses is our function, *h*, which is cos(*x*).Let’s plug that in. *f`(x)*=3(cos(*x*))^2 times the derivative of the inside function, cos(*x*).

Well I know that the derivative of cos(*x*) is -sin(*x*), so I can plug that in and simplify this equation to get *f`(x)*= -3(sin(*x*)cos^2(*x*).

## Fourth Example

Let’s do one more example. Let’s say *f(x)*=*x*^2 times the natural log (ln) of *x*. Well in this case, *x*^2 is like one function of *x* and the natural log of *x* is another, so I know I should use the product rule.

The product rule says that if you’re looking for the derivative of *u* times *v* that are both functions of *x*, the derivative is *uv`* + *vu`*, which is the first times the derivative of the second, plus the second times the derivative of the first. In our case, the first is *x*^2 and the second is ln(*x*). So let’s plug our functions into this rule.

The first, *x*^2, times the derivative (*d/dx*) of the second, ln(*x*), plus the second, ln(*x*), times the derivative of the first, *d/dx(x^2)*. I’m going to use my rules for the derivative of ln(*x*) being 1/*x*, and the derivative of *x^n* is *nx*^(*n*-1) and find that *x*^2*d/dx*(ln(*x*)) is *x*^2(1/*x*)). My second term we said was ln(*x*)*d/dx(x*^2); I get ln(*x*)2*x*.

Let’s simplify.

We’ll knock out this *x* and one of these 2*x*s, and let’s put this 2*x* on the left side of ln(*x*). So *f`x*=*x* + 2*xln(*x*). You could leave it like this, but in general, you’re going to want to simplify this a little further by factoring out an *x*. There’s an *x* in both terms, so I’m going to divide both terms by *x* and put it on the outside, so I have *x*(1 + 2ln(*x*)).*

## Lesson Summary

Let’s review.

There are a few formulas that you need to remember for calculating the derivatives. You should know what the derivative is of a constant, of a power like *x*^*n*, of any kind of exponential like *e*^*x*, or log like the natural log of *x*, and you should know some of the common trig derivatives, such as sin(*x*), cos(*x*) and tan(*x*). Once you get those formulas down, you should always remember the three most important derivative rules. One is the **product rule**, the second is the **quotient rule** and the third is the **chain rule**.

But the absolute most important thing that you can remember about derivatives, that will always help you when you’re thinking about derivatives, is that the derivative is a rate of change. So the derivative is the slope of the tangent to your function.