Reading: Derivatives Formulas
In this section, we’ll get the derivative rules that will let us find formulas for derivatives when our function comes to us as a formula. This is a very algebraic section, and you should get lots of practice. When you tell someone you have studied calculus, this is the one skill they will expect you to have. There’s not a lot of deep meaning here—these are strictly algebraic rules.Building Blocks
These are the simplest rules—rules for the basic functions. We won’t prove these rules; we’ll just use them. But first, let’s look at a few so that we can see they make sense.Example
Find the derivative of y = f(x) = mx + bSolution
This is a linear function, so its graph is its own tangent line! The slope of the tangent line, the derivative, is the slope of the line: f′(x) = mRule
The derivative of a linear function is its slope.Example
Find the derivative of f(x) = 135Solution
Think about this one graphically, too. The graph of f(x) is a horizontal line. So its slope is zero. f′(x) = 0Rule
The derivative of a constant is zero.Example
I will just tell you that the derivative of f(x) = x3 is f′(x) = 3x2. Now think about the function g(x) = 4x3. What will its derivative be?Solution
Think about what this change means to the graph of g—it’s now 4 times as tall as the graph of f. If we find the slope of a secant line, it will be [latex] \frac{\Delta g}{\Delta x} = \frac{4 \Delta f}{\Delta x} = 4 \frac{\Delta f}{\Delta x} [/latex]; each slope will be 4 times the slope of the secant line on the f graph. This property will hold for the slopes of tangent lines, too: [latex] \frac{d}{dx}(4x^3) = 4\frac{d}{dx}(x^3) = 4(3x^2) = 12x^2 [/latex]Rule
Constants come along for the ride. OK, enough of that. Here are the basic rules, all in oneDerivative Rules: Building Blocks
In what follows, f and g are differentiable functions of x.- Constant Multiple Rule: [latex] \frac{d}{dx} (kf) = kf\prime [/latex]
- Sum (or Difference) Rule: [latex] \frac{d}{dx}(f+g)=f\prime + g\prime [/latex] or [latex] \frac{d}{dx}(f-g)=f\prime - g\prime [/latex]
- Power Rule: [latex] \frac{d}{dx}(x^n) = nx^{n-1} [/latex] Special cases: [latex] \frac{d}{dx}(k) = 0 [/latex] (because k = kx0) and [latex] \frac{d}{dx}(x) = 1 [/latex] (because x = x1)
- Exponential Functions: [latex] \frac{d}{dx}(e^x) = e^x [/latex] [latex] \frac{d}{dx}(a^x) = \text{ln} a \times a^x [/latex]
- Natural Logarithm: [latex] \frac{d}{dx}(\text{ln}x) = \frac{1}{x} [/latex]
Example
Find the derivative of p(x) = 17x10 +13x8 1.8x + 1003Solution
[latex-display] \frac{d}{dx}(17x^{10} + 13x^8 - 1.8x + 10003) [/latex-display] [latex-display] = \frac{d}{dx}(17x^{10}) +\frac{d}{dx}(13x^8) -\frac{d}{dx}(1.8x) +\frac{d}{dx}(10003) [/latex-display] [latex-display] = 17\frac{d}{dx}(x^{10}) + 13\frac{d}{dx}(x^8) -\frac{d}{dx}(1.8x) +\frac{d}{dx}(10003) [/latex-display] [latex-display] =17(10x^9)+13(8x^7)-1.8(1)+0[/latex-display] [latex-display] = 170x^9 + 104x^7 - 1.8 [/latex-display] No, you don’t have to show every single step. Do be careful when you’re first working with the rules, but pretty soon you’ll be able to just write down the derivative directly.Example
[latex-display] \frac{d}{dx}(17x^2 - 33x + 12) = 34x - 33 [/latex-display] The power rule works even if the power is negative or a fraction. In order to apply it, first translate all roots and one-overs into exponents.Example
Find the derivative of [latex] y = 3\sqrt t - \frac{4}{t^4} + 5e^t [/latex]Solution
First step—translate into exponents: [latex] y = 3\sqrt t - \frac{4}{t^4} + 5e^t = 3t^{1/2} - 4t^{-4} + 5e^t [/latex] Now you can take the derivative: [latex-display] \frac{d}{dt}(3\sqrt t - \frac{4}{t^4} + 5e^t) = \frac{d}{dt}(3t^{1/2} - 4t^{-4} + 5e^t) [/latex-display] [latex-display] = 3(\frac{1}{2}t^{-1/2}) - 4(-4t^{-5}) + 5(e^t) = \frac{3}{2}t^{-1/2} + 16t^{-5} + 5e^t [/latex-display] Be careful when finding the derivatives with negative exponents. This video provides another example of finding the derivative of a function containing radicals. https://youtu.be/OqsVPz8I-_YExample
The cost to produce x items is [latex] \sqrt x [/latex] hundred dollars.- What is the cost for 100 items? 101 items? What is cost of the 101st item?
- For [latex] f(x) = \sqrt x [/latex], calculate f′(x) and evaluate f′ at x = 100. How does f′(100) compare with the last answer in part 1?
Solution
- Put [latex] f(x) = \sqrt x = x^{1/2} [/latex] hundred dollars, the cost for x items. Then f(100) = $1000 and f(101) = $1004.99, so it costs $4.99 for that 101st item. Using this definition, the marginal cost is $4.99.
- [latex] f\prime(x) = \frac{1}{2}x^{-1/2} [/latex] so [latex] f\prime(100) = \frac{1}{2 \sqrt {100}} = \frac{1}{20} = [/latex] hundred dollars = $5.00. Note how close these answers are! This shows (again) why it’s OK that we use both definitions for marginal cost.
Product and Quotient Rules
The basic rules will let us tackle simple functions. But what happens if we need the derivative of a combination of these functions?Example
Find the derivative of [latex] g(x) = (4x^3 - 11)(x + 3) [/latex]Solution
This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We can simply multiply it out to find its derivative: [latex-display] g(x) = (4x^3 - 11)(x + 3) = 4x^4 - 11x + 12x^3 - 33 [/latex-display] [latex-display] g\prime(x) = 16x^3 - 11 + 36x [/latex-display]Example
Find the derivative of [latex] f(x) = (4x^5 + x^3 - 1.5x^2 - 11)(x^7 - 7.25x^5 + 120x + 3) [/latex]Solution
This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We could simply multiply it out to find its derivative as before – who wants to volunteer? Nobody? We’ll need a rule for finding the derivative of a product so we don’t have to multiply everything out. Is the rule what we hope it is, that we can just take the derivatives of the factors and multiply them? Unfortunately, no—that won’t give the right answer.Example
Find the derivative of [latex] g(x) = (4x^3 - 11)(x+3) [/latex]Solution
We already worked out the derivative. It’s [latex] g\prime(x) = (4x^3 - 11)(x + 3) [/latex]. What if we try differentiating the factors and multiplying them? We’d get [latex] (12x^2)(1) = 12x^2 [/latex], which is totally different from the correct answer. The rules for finding derivatives of products and quotients are a little complicated, but they save us the much more complicated algebra we might face if we were to try to multiply things out. They also let us deal with products where the factors are not polynomials. We can use these rules, together with the basic rules, to find derivatives of many complicated looking functions.Derivative Rules: Product and Quotient Rules
In what follows, f and g are differentiable functions of x.- Product Rule: [latex] \frac{d}{dx} (fg)=f\prime g + fg\prime [/latex] The derivative of the first factor times the second left alone, plus the first left alone times the derivative of the second. The product rule can extend to a product of several functions; the pattern continues—take the derivative of each factor in turn, multiplied by all the other factors left alone, and add them up.
- Quotient Rule: [latex] \frac{d}{dx}(\frac{f}{g})= \frac{f\prime g - fg\prime}{g^2} [/latex] The numerator of the result resembles the product rule, but there is a minus instead of a plus; the minus sign goes with the g′. The denominator is simply the square of the original denominator—no derivatives there.
Example
Find the derivative of [latex] F(t)= e^t \text{ln}t [/latex]Solution
This is a product, so we need to use the product rule. I like to put down empty parentheses to remind myself of the pattern; that way I don’t forget anything. [latex-display] F\prime(t)=()() + ()() [/latex-display] Then I fill in the parentheses—the first set gets the derivative of et, the second gets left alone, the third gets left alone, and the fourth gets the derivative of lnt. [latex-display] F\prime(t)=(e^t)(\text{ln}t) + (e^t)(\frac{1}{t}) [/latex-display] Notice that this was one we couldn’t have done by “multiplying out.”Example
Find the derivative of [latex] y = \frac {x^4 + 4^x}{3 + 16x^3} [/latex]Solution
This is a quotient, so we need to use the quotient rule. Again, I find it helpful to put down the empty parentheses as a template: [latex-display] y\prime = \frac {()()-()()}{()^2} [/latex-display] Then I fill in all the pieces: [latex-display] y\prime = \frac {(4x^3 + \text{ln}4 \times 4^x)(3 + 16x^3)-(x^4 + 4^x)(48x^2)}{(3+16^3)^2} [/latex-display] Now for goodness’ sakes don’t try to simplify that! Remember that “simple” depends on what you will do next; in this case, we were asked to find the derivative, and we’ve done that. Please STOP!Chain Rule
There is one more type of complicated function that we will want to know how to differentiate: composition. The Chain Rule will let us find the derivative of a composition. (This is the last derivative rule we will learn!)Example
Find the derivative of [latex] y=(4x^3 + 15x)^2 [/latex].Solution
This is not a simple polynomial, so we can’t use the basic building block rules yet. It is a product, so we could write it as [latex] y=(4x^3 + 15x)^2 =(4x^3 + 15x)(4x^3 + 15x) [/latex] and use the product rule. Or we could multiply it out and simply differentiate the resulting polynomial. I’ll do it the second way: [latex-display] y=(4x^3 + 15x)^2 = 16x^6 +120x^4 + 225x^2 [/latex-display] [latex-display] y=64x^5 + 480x^3 +450x [/latex-display]Example
Find the derivative of [latex] y=(4x^3 + 15x)^20 [/latex].Solution
We could write it as a product with 20 factors and use the product rule, or we could multiply it out. But I don’t want to do that, do you? We need an easier way, a rule that will handle a composition like this. The Chain Rule is a little complicated, but it saves us the much more complicated algebra of multiplying something like this out. It will also handle compositions where it wouldn’t be possible to “multiply it out.” The Chain Rule is the most common place for students to make mistakes. Part of the reason is that the notation takes a little getting used to. And part of the reason is that students often forget to use it when they should. When should you use the Chain Rule? Almost every time you take a derivative.Derivative Rules: Chain Rule
In what follows, f and g are differentiable functions with [latex] y=f(u) [/latex] and [latex] u=g(x) [/latex]Chain Rule (Leibniz notation)
[latex-display] \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} [/latex-display] Notice that the dus seem to cancel. This is one advantage of the Leibniz notation; it can remind you of how the chain rule chains together.Chain Rule (using prime notation)
[latex-display] f\prime (x) = f\prime (u) \times g\prime (g(x)) \times g\prime(x) [/latex-display]Chain Rule (in words)
The derivative of a composition is) the derivative of the outside TIMES the derivative of what’s inside. I recite the version in words each time I take a derivative, especially if the function is complicated.Example
Find the derivative of [latex] y= (4x^3 + 15x)^2 [/latex].Solution
This is the same one we did before by multiplying out. This time, let’s use the Chain Rule: The inside function is what appears inside the parentheses: [latex] 4x^3 + 15x [/latex]. The outside function is the first thing we find as we come in from the outside—it’s the square function, something2. We want the derivative of the outside (2 something) TIMES the derivative of what’s inside (which is [latex] 12x^2 + 15 [/latex]): [latex-display] y= (4x^3 + 15x)^2 [/latex-display] [latex-display] y\prime = 2(4x^3 + 15x)(12x^2 + 15) [/latex-display] (By the way, if you multiply this out, you get the same answer we got before. Hurray! Algebra works!)Example
Find the derivative of [latex] y= (4x^3 + 15x)^{20} [/latex]Solution
Now we have a way to handle this one. It’s the derivative of the outside TIMES the derivative of what’s inside. [latex-display] y= (4x^3 + 15x)^{20} [/latex-display] [latex-display] y\prime = 20(4x^3 + 15x)^{19} \times (12x^2 + 15) [/latex-display]Example
Differentiate [latex] e^{x^2 + 5} [/latex].Solution
This isn’t a simple exponential function; it’s a composition. Typical calculator or computer syntax can help you see what the “inside” function is here. On a TI calculator, for example, when you push the ex key, it opens up parentheses: ex2+ 5. Here, the inside is the exponent. Now we can use the Chain Rule: We want the derivative of the outside TIMES the derivative of what’s inside. The outside is the “e to the” function, so its derivative is the same thing. The derivative of what’s inside is 2x. So [latex-display] \frac{d}{dx}(e^{x^2 + 5}) = (e^{x^2 + 5})(2x) [/latex-display]Example
The table gives values for f, f′, g, and g′ at a number of points. Use these values to determine (f°g)(x) and (f°g)′(x) at x = −1 and 0.| x | f(x) | g(x) | f′(x) | g′(x) | (f°g)(x) | (f°g)′(x) |
| –1 | 2 | 3 | 1 | 0 | ||
| 0 | –1 | 1 | 3 | 2 | ||
| 1 | 1 | 0 | –1 | 3 | ||
| 2 | 3 | –1 | 0 | 1 | ||
| 3 | 0 | 2 | 2 | –1 |
Solution
( f°g )(–1) = f( g(–1) ) = f( 3 ) = 0 and ( f°g )(0) = f( g(0) ) = f( 1 ) = 1. ( f°g )′(–1) = f′( g(–1) ).g′( –1 ) = f′( 3 ).(0) = (2)(0) = 0 and ( f°g )′( 0 ) = f′( g( 0 ) ).g′( 0 ) = f′( 1 ).( 2 ) = (–1)(2) = –2 . I’ll let you do the rest.Derivatives of Complicated Functions
You’re now ready to take the derivative of some mighty complicated functions. But how do you tell what rule applies first? Come in from the outside–what do you encounter first? That’s the first rule you need. Use the Product, Quotient, and Chain Rules to peel off the layers, one at a time, until you’re all the way inside.Example
Find [latex] \frac{d}{dx}(e^{3x} \times \text{ln}(5x+7)) [/latex]Solution
Coming in from the outside, I see that this is a product of two (complicated) functions. So I’ll need the Product Rule first. I’ll fill in the pieces I know, and then I can figure the rest as separate steps and substitute in at the end: [latex-display] \frac{d}{dx}(e^{3x} \times \text{ln}(5x+7)) = (\frac{d}{dx}(e^{3x}))(\text{ln}(5x+7)) + (e^{3x})(\frac{d}{dx}(\text {ln}(5x + 7))) [/latex-display] Now as separate steps, I’ll find [latex] \frac{d}{dx}(e^{3x}) =3e^{3x} [/latex] (using the Chain Rule) and [latex] \frac{d}{dx}(\text{ln}(5x+7)) = \frac {1}{5x + 7} \times 5 [/latex] (also using the Chain Rule). Finally, to substitute these in their places: [latex-display] \frac{d}{dx}(e^{3x} \times \text{ln}(5x+7)) = (3e^{3x})(\text{ln}(5x+7)) + (e^{3x}) (\frac {1}{5x + 7} \times 5) [/latex-display] (And please don’t try to simplify that!)Example
Differentiate [latex] z = (\frac {3t^3}{e^t(t-1)})^4 [/latex]Solution
Don’t panic! As you come in from the outside, what’s the first thing you encounter? It’s that 4th power. That tells you that this is a composition, a (complicated) function raised to the 4th power. Step One: Use the Chain Rule. The derivative of the outside TIMES the derivative of what’s inside. [latex-display] \frac{dz}{dt}=\frac{d}{dt}(\frac{3t^3}{e^t(t-1)})^4 = 4(\frac{3t^3}{e^t(t-1)})^3 \times \frac {d}{dt}(\frac{3t^3}{e^t(t-1))}) [/latex-display] Now we’re one step inside, and we can concentrate on just the [latex] \frac {d}{dt}(\frac{3t^3}{e^t(t-1)}) [/latex] part. Now, as you come in from the outside, the first thing you encounter is a quotient—this is the quotient of two (complicated) functions. Step Two: Use the Quotient Rule: [latex-display] \frac {d}{dt}(\frac{3t^3}{e^t(t-1)}) = \frac{(9t^2)(e^t(t-1)) - (3t^3)(\frac{d}{dt}(e^t(t-1)))}{(e^t(t-1)^2)} [/latex-display] Now we’ve gone one more step inside, and we can concentrate on just the [latex] \frac{d}{dt}(e^t(t-1)) [/latex] part. Now we have a product. Step Three: Use the Product Rule: [latex-display] \frac{d}{dt}(e^t(t-1)) = (e^t)(t-1) + (e^t)(1) [/latex-display] And now we’re all the way in—no more derivatives to take. Step Four: Now it’s just a question of substituting back—be careful now! [latex] \frac{d}{dt}(e^t(t-1)) = (e^t)(t-1) + (e^t)(1) [/latex], so [latex] \frac{d}{dt}(\frac{3t^3}{e^t(t-1)}) = \frac{(9t^2)(e^t(t-1)) - (3t^3)((e^t)(t-1)+(e^t)(1))}{(e^t(t-1)^2)} [/latex], so [latex-display] \frac{dz}{dt} = \frac {d}{dt}(\frac{3t^3}{e^t(t-1)})^4 = 4(\frac{3t^3}{e^t(t-1)})^3 (\frac{(9t^2)(e^t(t-1)) - (3t^3)((e^t)(t-1)+(e^t)(1))}{(e^t(t-1)^2)}) [/latex-display] Phew!Licenses & Attributions
CC licensed content, Shared previously
- Business Calculus. Provided by: Washington State Colleges Authored by: Dale Hoffman and Shana Calaway. Located at: https://docs.google.com/file/d/0B1lkHWwO61QEM0gwOFhES2N5Tlk/edit. License: CC BY: Attribution.
- Power Rule. Authored by: James Sousa. License: CC BY: Attribution.
- Ex: Find the Derivative of a Function Containing Radicals. Authored by: James Sousa. License: CC BY: Attribution.
