# Applying the Rules of Differentiation to Calculate Derivatives Video

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Lesson Transcript
Instructor: Zach Pino
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^nth power. The derivative of x^nth 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 2x^(2 - 1). Well, 2 - 1 is 1, so this is 2x^1, or just 2x. So I can plug that in, and f`(x)=(x^2(cos(x)) - sin(x)(2x)) / (x^2)(x^2).

Let's simplify. Let's move this 2x 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(xcos(x) - 2sin(x)). Let's cancel this x with one of these four xs that we have on the bottom. With all of that, I find that the derivative of f(x) is (xcos(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 2x.

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 2x. So f`(x)=2xe^(x^2)

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