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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 29: Derivatives capstone

# Derivative rules review

Review all the common derivative rules (including Power, Product, and Chain rules).

## Basic differentiation rules

Constant rule: $\frac{d}{dx}\left(k\right)=0$
Sum rule: $\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]={f}^{\prime }\left(x\right)+{g}^{\prime }\left(x\right)$
Difference rule: $\frac{d}{dx}\left[f\left(x\right)-g\left(x\right)\right]={f}^{\prime }\left(x\right)-{g}^{\prime }\left(x\right)$
Constant multiple rule: $\frac{d}{dx}\left[kf\left(x\right)\right]=k{f}^{\prime }\left(x\right)$

## Power rule

$\frac{d}{dx}\left({x}^{n}\right)=n\cdot {x}^{n-1}$

## Product rule

$\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]={f}^{\prime }\left(x\right)g\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)$

## Quotient rule

$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{\left[g\left(x\right){\right]}^{2}}$

## Chain rule

$\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right)$

## Want to join the conversation?

• Can somebody explain the chain rule in an "easy-to-understand" way? Thanks :)
• Consider a function (x+1)^2.
We can see that the function has two parts, the enclosing part
(outside: ( )^2) and the enclosed part (inside: x+1).

To do the chain rule you first take the derivative of the outside as if you would normally (disregarding the inner parts), then you add the inside back into the derivative of the outside.
Afterwards, you take the derivative of the inside part and multiply that with the part you found previously.

So to continue the example:
d/dx[(x+1)^2]
1. Find the derivative of the outside:
Consider the outside ( )^2 as x^2 and find the derivative
as d/dx x^2 = 2x
the outside portion = 2( )
2. Add the inside into the parenthesis:
2( ) = 2(x+1)
3. Find the derivative of the inside and multiply:
as d/dx [x+1] = 1
1*2(x+1) = 2(x+1).

Thus, d/dx[(x+1)^2] = 2(x+1)
• The variable power rule is missing: a^x = ln(a) a^x and differentiation rules for logarithms
• Those are just special cases. These are the general overarching rules which apply to any old generic function f(x).
• In terms of the AP exam: Are proofs even needed? I know that you need to get a good conceptual idea, but my math teacher claims no one ever uses proofs. If they want, they can take a course on proofs in college, but teachers would lose kids interest with all these proofs. I'm 99 percent sure proofs are not required on the AP exam? Someone correct me if I'm wrong.
• Proofs are NOT specifically needed for the AP exam. However, working through proofs can be illuminating into how functions (and their derivations) work. Practically speaking, the more you are able to manipulate functions algebraically and trigonometrically, the better you will be able to work with function problems at the AP (or any) exam. It's a matter of familiarity, fluency, and functionality. You don't have to be able to do a proof at the exam, but if you've done a lot of proofs beforehand, you'll likely score better at the exam.
• In quotient rule, can't we write the derivative of f(x)/g(x) as
f(x).g'(x)^-1 + g(x)^-1.f'(x) ?
• No, because d/dx (1/g(x)) is not 1/g'(x). You have to apply the chain rule. It should be -1/(g(x))² · g'(x)
So we'd have (using the product rule and the chain rule):
d/dx f(x) · 1/g(x) = 1/g(x) · f'(x) + f(x) · -1/(g(x))² · g'(x)

Which, with a bit of manipulation, can be made to look like the familiar quotient rule for differentiation.
• Can you apply chainrule for questions like: ddx of [f(g(h(x)))]?
• Yes, you would use the chain rule twice for that case.
• we could also use the power rule to prove that d/dx[x]=1
• Yes, you could, x is the same thing as x^1, so it becomes 1x^0 when you apply the power rule, which is the same thing as 1
• how do you derive the square root of x to the 3rd power
• For the equation y = x(x - 4) ^ 3 you have to use the product rule AND the chain rule since it's x * (x-4)^3. If you're using the formula d/dx f(x)g(x) = f'(x)g(x) + g'(x)f(x) then your f(x)=x and g(x)=(x-4)^3 so f'(x)=1 and g'(x)=3(x-4)^2, so if you plug it in you get 1 * (x-4)^3 + 3(x-4)^2 * x which I simplified to (x - 4)^2 (x-4+3x) or (x-4)^2(4x-4).