Question

Find the derivative by the limit process

Answer 1

\[f(x)=x^{3}+x^{2}\]

Answer 2

\(\large f'(x) = \lim \limits_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

Answer 3

o.o

Answer 4

\(\large f'(x) = \lim \limits_{h \to 0} \frac{(x+h)^3+(x+h)^2 - (x^3+x^2)}{h}\)

Answer 5

simplify

Answer 6

why are you using h instead of a

Answer 7

good question :)
there are two definitions actually.

if you're comfortable wid a, lets use that oly :)

Answer 8

Thank you!

Answer 9

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)

Answer 10

Yes, dont forget the bottom x-a

Answer 11

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)


\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 - a^3-a^2}{x-a}\)

Answer 12

\[\lim_{x \rightarrow a} (x ^{3}+x ^{2})-(a ^{3}+a ^{2})\div(x-a)\]

Answer 13

you can use \frac{}{} to show it in bottom :)

Answer 14

why do you say f ' (x) not f(x)

Answer 15

f(x) is the function,
f'(x) is the \(derivative \) of funciton f(x)

Answer 16

\[\frac{\lim_{x \rightarrow a} (x ^{3}+x ^{2})-(a ^{3}+a ^{2}) }{ (x-a)}\]

Answer 17

So I should but f'(x) not lim x-->a

Answer 18

looks u nailed it ! just put the limit out of frac :)

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)


\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 - a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3- a^3) + (x^2-a^2)}{x-a}\)

Answer 19

or that's in front of limx-->a

Answer 20

its just a notational thing :-


if \(f(x)\) is the given funciton,
we say \(f'(x)\) is the derivative of given function \(f(x)\)

Answer 21

and the derivative \(f'(x)\) is defined as :-

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)

Answer 22

so you should put both f'(x) and lim x->a

Answer 23

\[\large f'(x) = \lim \limits_{x \to a} \frac{(x- a)^3 + (x-a)^2}{x-a}\]

Answer 24

no, you're mixing both the definitions. i shouldnt have shown u the 2nd definition :o

Answer 25

Sorry :'(

Answer 26

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)


\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) - (a^3+a^2)}{x-a}\)

Answer 27

that's it?

Answer 28

no, you need to take the limit and find the derivative still

Answer 29

to take the limit, you need to put x = a

Answer 30

if u put x = a, wat happens to the bottom expression ?

Answer 31

0

Answer 32

bottom u have \(x-a\)
if u put x = a, bottom becomes \(a-a = 0\)

Answer 33

Yes ! So bottom becomes the nice number \(0\)
which is illegal in math

Answer 34

it is illegal to have \(0\) in the denominator in math ok

Answer 35

call da cops

Answer 36

lol not yet, this is a friendly funciton... we dont call cops when friends act tough ;)

Answer 37

so we need to mess wid the numerator, and see if we can cancel out denominator somehow

Answer 38

let me show u the complete solution once, then it wil make more sense to you

Answer 39

ok

Answer 40

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) - (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 - a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3-a^3 + x^2-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3-a^3) + (x^2-a^2)}{x-a}\)

Answer 41

you should be okays so far ?

Answer 42

next, you need to pull two identities :-
\(a^3-b^3 = (a-b)(a^2+ab+b^2)\)

\(a^2-b^2= (a+b)(a-b)\)

Answer 43

YES

Answer 44

seen them before ?

Answer 45

NO

Answer 46

wait it's like factoring cube and square roots in algebra

Answer 47

right?

Answer 48

yes, they're the identities/formulas.
you should be like memorizing these.... so that u can use these at will

Answer 49

ok

Answer 50

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) - f(a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) - (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 - a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3-a^3 + x^2-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3-a^3) + (x^2-a^2)}{x-a}\)

plug the identities...

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2) + (x-a)(x+a)}{x-a}\)

Answer 51

you should be excited nw seeing (x-a) s on the numerators also... we're going to cancel denominator wid them !

Answer 52

In the third step why do u put a^3-a^2, and in step 2 it is +

Answer 53

\(\large x^3+x^2 - a^3-a^2 \)


ive just rearranged the terms
cubes separately, squares separately...

Answer 54

so that we can apply the known identities

Answer 55

\(\color{Red}{x^3}+x^2 \color{red}{-a^3}-a^2\)

Answer 56

OMG sorry lol

Answer 57

np :) good to make sure u *get* it

Answer 58

let me knw once u digest above steps... :)

Answer 59

digested. So when I cancel x-a from the denominator do I cancel both x-a numerators or just one?

Answer 60

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) – f(a)}{x-a} \)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) – (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 – a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3-a^3 + x^2-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3-a^3) + (x^2-a^2)}{x-a}\)

plug the identities...

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2) \color{red}{+} (x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)}{x-a} \color{red}{+} \frac{(x-a)(x+a)}{x-a}\)

Answer 61

So... ? both are getting cancelled out eh ?

Answer 62

yes

Answer 63

\[f(x)=\lim_{x \rightarrow a}(x^2+ax+a^2)(x+a)\]

Answer 64

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) – f(a)}{x-a} \)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) – (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 – a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3-a^3 + x^2-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3-a^3) + (x^2-a^2)}{x-a}\)

plug the identities...

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2) \color{red}{+} (x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)}{x-a} \color{red}{+} \frac{(x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} ~~~ x^2+ax+a^2\color{red}{+} x+a\)

Answer 65

:(

Answer 66

its a + sign between them ok

Answer 67

its ok :) you're doing it first time... and you're doing great ok

Answer 68

ok so if there any more simplifying to do

Answer 69

is*

Answer 70

now that denominator is gone, simply put x = a

Answer 71

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) – f(a)}{x-a} \)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) – (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 – a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3-a^3 + x^2-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3-a^3) + (x^2-a^2)}{x-a}\)

plug the identities...

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2) \color{red}{+} (x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)}{x-a} \color{red}{+} \frac{(x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} ~~~ x^2+ax+a^2\color{red}{+} x+a\)

\(\large f'(x) = a^2+a*a+a^2\color{red}{+} a+a\)

Answer 72

simpify, we're done.

Answer 73

\[\large f'(x) = \lim \limits_{x \to a} ~~~ a^2+2a+a^2\color{red}{+} a+a\]

Answer 74

\[2a^2+4a\]

Answer 75

hey one imp thing,
after pluggin in x = a, you MUST NOT show lim x->a

Answer 76

Sorry!

Answer 77

\(\large f'(x) = \lim \limits_{x \to a} \frac{f(x) – f(a)}{x-a} \)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) – (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3+x^2) – (a^3+a^2)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3+x^2 – a^3-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{x^3-a^3 + x^2-a^2}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x^3-a^3) + (x^2-a^2)}{x-a}\)

plug the identities...

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2) \color{red}{+} (x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)}{x-a} \color{red}{+} \frac{(x-a)(x+a)}{x-a}\)

\(\large f'(x) = \lim \limits_{x \to a} ~~~ x^2+ax+a^2\color{red}{+} x+a\)

\(\large f'(x) = a^2+a*a+a^2\color{red}{+} a+a\)

\(\large f'(x) = a^2+a^2+a^2\color{red}{+} a+a\)

\(\large f'(x) = 3a^2\color{red}{+} 2a\)

\(\large f'(x) = 3x^2\color{red}{+} 2x\)

Answer 78

the derivative of x^3 + x^2 is 3x^2+2x

Answer 79

see if that makes some sense

Answer 80

how?

Answer 81

how for what

Answer 82

how can i check the answer

Answer 83

oh sorry this time we cannot check the answer... as the given function is not a simple line...

Answer 84

for line, derivative is the slope of line...
but for other functions, its not possible to generalize derivative like that