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
Some of the most commonly used 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!
Commonly used trig functions
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.
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.
Finding the derivative in example #1
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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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).
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^2d/dx(ln(x)) is x^2(1/x)). My second term we said was ln(x)d/dx(x^2); I get ln(x)2x.
The product rule is used in example #4
Let's simplify. We'll knock out this x and one of these 2xs, and let's put this 2x on the left side of ln(x). So f`x=x + 2xln(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)).
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.
There are many rules that help us take the derivative of any function we are provided with. Here are some more practice problems where you have to take derivatives. Solutions are provided under the Solutions section.
1. Take the derivative of y = cos(2x)
2. Take the derivative of y = 2x2 / (x - 1)
3. Take the derivative of y = (x - 1)(x2)
4. Take the derivative of y = ex
5. Take the derivative of y = ln(x)
1. Derivative of trigonometric function and chain rule.
y' = -sin(2x)(2)
y' = -2sin(2x)
2. Quotient rule.
y' = u'v - uv' / v2
y' = (4x)(x - 1) - (2x2)(1) / (x2 - 2x + 1)
y' = 4x2 - 4x - 2x2 / (x2 - 2x + 1)
y' = 2x2 - 4x / (x2 - 2x + 1)
3. Product rule.
y' = u'v + uv'
y' = (1)(x2) + (x - 1)(2x)
y' = x2 + 2x2 - 2x
y' = 3x2 - 2x
4. Special case derivative.
y' = ex
5. Special case derivative.
y' = 1/x
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