Question

Note: This is NOT a question. This is a tutorial

How to take the derivative of a function? See comment below for a tutorial on the basic rules on derivatives!

Answer 1

Differential Calculus Basics.

Taking the derivative of a function is easy, if you know what to do. There are some rules in derivatives. (Note: these rules are not the official names. I am calling them names based on how they are used for the convenience of the tutorial)

Rule #1: Constant Rule. The derivative of a constant (without any variables) is always zero. This means that the derivative of ALL positive, negative, and zero integers (i.e. -1, 1, 0) are all zero. Note that these numbers must NOT contain any variables so that their derivative will be zero. If they contain variables, then it is under another rule.

Rule #2: Single Variable Rule. The derivative of a single variable with 1 as the exponent is always 1. (i.e. derivative of x = 1).

Rule #3: Constant Multiple Rule. To take the derivative of the of a constant containing a variable, factor out the constant, then take the derivative of the variable. For example, to take the derivative of 3x, factor out 3 so it will be 3(x). Now take the derivative of the variable inside, which in this case is x. So it will be 3(derivative of x). From the previous rule, we know that the derivative of x is 1 so it will become 3(1) = 3.

Rule #4: Power Rule. To take the derivative of a variable with an exponent higher than one, first, multiply the exponent by the variable. Then, subtract 1 from the exponent. For example, to take the derivative of \(x^2\), first multiply the exponent (which in this case is 2) to the variable. So, we have \(2x^2\). Now, subtract 1 from the exponent, so it will become \(2x^{2-1} = 2x^1 = 2x\)

Important note: When given an expression, we take the derivative one-by-one each term. For example, we are given \(5x^2 - 3x + 1\). To take the derivative of this, we take the derivative of \(5x^2\), add it to the derivative of \(-3x\), and add that to the derivative of \(1\). What is the derivative of \(5x^2\)? First, we use the constant multiple rule. We pull out the constant. \(5(x^2)\). Now, we take the derivative of \(x^2\). How? We use the power rule. Multiply the exponent by the variable, then subtract the exponent by 1. So, we'll have \(5(2x^{2-1}) = 5(2x^1) = 5(2x) = 10x\). This is now the derivative of the first term. Now, we take the derivative of the second term. We'll use the constant multiple rule. Pull out the constant, then take the derivative of the variable, so we'll have \(-3(derivative of x) = -3(1) = -3\). Lastly, we take the derivative of the last term. We use the Constant Rule to find out that the derivative of the last term is 0. So, the derivative of our expression is \(10x - 3 +0 = 10x - 3\)

Answer 2

@GOODMAN

Answer 3

Woahh, you did this all now?

Answer 4

yepppp :DDD like itttt?

Answer 5

@lgbasallote shuper speed! of writing!

Answer 6

thanks @Rohangrr :)

Answer 7

@lgbasallote u type more faster thn me!

Answer 8

@lgbasallote you did awesome, wow, i might actually pass next year :DDDD

Answer 9

lol i typed that in notepad first :P then i copy-pasted. I wrote it for 30 minutes

Answer 10

im glad this could help goodie :DDD

Answer 11

note: i did not mention some of the other rules here as they are not really as basic as the ones stated

Answer 12

Basic is all i wanted, and this is great!! Thanks LG :D

Answer 13

Here are some general forms for your viewing pleasure.\[\frac{d}{dx}f(x)=f'(x)\]\[\frac{d}{dx}f(x)g(x)=f'(x)g(x)+g'(x)f(x)\]\[\frac{d}{dx}(f\circ g)(x)=\frac{d}{dx}f(g(x))=g'(x)f'(g(x))\]\[\frac{d}{dx}\left(f(x)+g(x)\right)=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)\]\[\frac{d}{dx}nf(x)=n\frac{d}{dx}f(x)\]\[\frac{d}{dx}x^n=nx^{n-1}\]\[\frac{d}{dx}\ln x=\frac{1}{x}\]\[\frac{d}{dx}\sin x=\cos x\]\[\frac{d}{dx}\cos x=-\sin x\]\[\frac{d}{dx}\tan x=\sec^2x\]

Answer 14

Here's the tricky question.\[\frac{d}{dx}n^x=\,?\]The smart thing to do:\[y=n^x\]\[\ln y=x\ln n\]Now, taking the derivative of both sides (and keeping in mind the chain rule)...\[y'\frac{1}{y}=\ln n\]\[y'=y\ln n\]\[\frac{dy}{dx}=n^x\ln n\]

Answer 15

you forgot good ol \(\large\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\) :O

By the way...dont mind these latter comments they are more advanced derivatives

Answer 16

\[\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{d}{dx}\left(f(x)g(x)^{-1}\right)\]Which is just product and chain rule. :D

Answer 17

Why force people to memorize another rule, when we already have so many?

Answer 18

hahaha =)) i just memorize the 4 i stated + product + quotient + chain..but i guess quotient rule can just be got from chain and product....anyway these are more intermediate derivatives.

Answer 19

and btw...you used implicit differentiation on that second to the last example no fair :P haha

Answer 20

Implicit differentiation is not different from regular differentiation. :D

Answer 21

Thank you lgbasallote and badreferences for the great stuff here.

Answer 22

lgbasallote needs a section for tutorials -- preferably for any math that's not 'special' arithmetic. :D

Answer 23

I'd be willing to help with that section.

Answer 24

hahahaha i think i know those "special" arithmetic :P =)))

Answer 25

@lgbasallote , not a question, you break CoC, bad girl with bunny xD

Answer 26

Nice job, btw

Answer 27

aww pwees dont weepowt meee

Answer 28

Amazing :D

Answer 29

that means so muchhh :""> thanks @lalaly

Answer 30

Nice work

Answer 31

I recently found this.Quite interesting http://betterexplained.com/articles/calculus-building-intuition-for-the-derivative/

Answer 32

@lgbasallote Great work. ;)

Answer 33

... lol I do know "Basic" integrals :P...
\[\LARGE \int x^3+2x^2 dx=\int x^3dx+2\int x^2dx=\]
\[\LARGE \frac{x^4}{4}+2\cdot \frac{x^3}{3}+C\]
is that enough ? lol hahha...
\[\LARGE \int x^ndx=\frac{x^{n+1}}{n+1}+C\]
...I'll check it out anyway.. ;)
I'm learning from khan academy it's great...
integrals by part... (because I know by substitution too) is just that I'm a bit lame on that.. :P

Answer 34

i think the post is still under construction but wait for it..

Answer 35

yeah.. ;) brb ...

Answer 36

Kahanga-hanga :-). Very nice initiative. I always like to refresh my Calc.

Answer 37

nice igbas..

Answer 38

thanks ruchi :) i worked hard for it :DD