Question

lim as dx--->0. (5(x+dx)-(x+dx)-(5x^2-x))/dx

Answer 1

how about infinity

Answer 2

simplify the numerator 1st...

Answer 3

well, i'd like to know the process, you know? like the distributing & cancelling out? Because i got to 5x^3+15x^2dx-15xdx^2+5dx^3-dx-5x^2 and idk if that's right or what to do next.

Answer 4

@akash123 is that simplification right? ^^^

Answer 5

let's do the simplification together..

Answer 6

ok thanks

Answer 7

1st two terms of the numerator...5(x+dx)-(x+dx)

Answer 8

5x+5 dx - x-dx
does it make sense?

Answer 9

5(x+dx)-(x+dx) = 5 * (x+dx)-1 * (x+dx)

Answer 10

i forgot to add that the (x+dx) is square.d, sorry. but yes that makes sense.

Answer 11

the first one.

Answer 12

ok ...write the complete question again..

Answer 13

(5(x+dx)^2 -(x+dx)-(5x^2-x))/dx

Answer 14

now it's ok?

Answer 15

(5(x+dx)^2 - (x+dx) - (5x^2-x))/dx

Answer 16

you know the identity (a+b)^2 = a^2 + b^2 + 2 a b

Answer 17

yessss sir

Answer 18

so using this identity 1st term of the numerator
5(x+dx)^2 = 5( x^2+ dx ^2 +2 x*dx)

Answer 19

I m not a sir...i m akash only

Answer 20

sorry I say that, i wasn't calling you a sir.
and yes, I'm following you.

Answer 21

5(x+dx)^2 -(x+dx)-(5x^2-x) = 5( x^2+ dx ^2 +2 x*dx) - x -dx - 5 x^2 +x

Answer 22

uh huh

Answer 23

5( x^2+ dx ^2 +2 x*dx) - x -dx - 5 x^2 +x = 5x^2+ 5dx ^2 + 10 x*dx- x -dx - 5 x^2 +x

Answer 24

now look for similar terms..

Answer 25

ohhhhhhh 5x^2 & Xs

Answer 26

yes..

Answer 27

5dx ^2 + 10 x*dx-dx

Answer 28

it's ok?

Answer 29

5x^2 - 5x^2+x -x ...got cancelled

Answer 30

yes, I realize the simplified version i posted corresponded to another problem, yes I got that. what's next?

Answer 31

so u have 5dx ^2 + 10 x*dx-dx in the numerator after simplification

Answer 32

now take dx common from numerator...

Answer 33

i was not underestimating u...if u think so...sorry

Answer 34

dx ( 5dx + 10 x-1)

Answer 35

5dx(x+2x-1)/ dx so you cancel the dx on top & bottom then you plug in?
it's fine, i didn't think that

Answer 36

ohhhh i see what you did.

Answer 37

after taking dx common u'll get dx ( 5dx + 10 x-1)

Answer 38

yeah why don't you take out the 5?

Answer 39

u can take also 5 common...u'll be left with dx 5( dx + 2 x-1/5)

Answer 40

ok so same thing?

Answer 41

nevermind. what's next?

Answer 42

cancel the dx from numerator and denominator...

Answer 43

and put dx=0

Answer 44

after canceling ( 5dx + 10 x-1) or 5( dx + 2 x-1/5)

Answer 45

one is with dx common and other is by taking common 5 dx

Answer 46

yes...

Answer 47

now plug in dx=0

Answer 48

-1?

Answer 49

no...10x-1

Answer 50

u have 10 x also in the expression...

Answer 51

ohhh ok. so what do you do with the 10x-1?

Answer 52

it's the final answer of your question

Answer 53

as u have been asked....lim as dx--->0

Answer 54

oh ok so you know you're at the final answer when there are no more dxs?

Answer 55

yes...u have been given an expression ...(5(x+dx)^2 - (x+dx) - (5x^2-x))/dx...and u are being asked...what will be value of the expression if dx--->0...so there should be no dx in ur final answer.

Answer 56

u can try this limit lim x---->2 (x^2-4)/(x-2)

Answer 57

lim x---->2 (x^2-4)/(x-2)

Answer 58

lim=4

Answer 59

yes...perfectly right...as u r seeing here there is no x in ur final answer.

Answer 60

similarly there wont be any dx in the answer of your question...and x will be treated as a constant..

Answer 61

oh ok. do you think you can help me with one more question?

Answer 62

yes...ask

Answer 63

sorry my computer went crazy on me

Answer 64

ok so it's 5x^3-10x^2+12

Answer 65

and I know you have to plug in x+dx in the xs

Answer 66

and I know you have to plug in x+dx in the xs
didnt get it?

Answer 67

like 5(x+dx)^3...

Answer 68

got it...u have to plug (x+dx) in the place of x everywhere and plug in dx=0..is it so?

Answer 69

i don't know about the dx =0 but after distributing the 5 and 10 I got

Answer 70

write down the complete question...

Answer 71

find lim as dx----> 0. (f(x +dx) - f(x))/dx. those are the directions.

Answer 72

actually this id definition of f'(x)

Answer 73

then the problem is 5x^3-10x^2-12

Answer 74

and u have to calculate f'(x) for a given f(x)

Answer 75

now i got the question..

Answer 76

here f(x) = 5x^3-10x^2-12

Answer 77

and u have to find f'(x) using
f'(x)= dx----> 0. (f(x +dx) - f(x))/dx

Answer 78

now 1st calculate f(x+dx)=5(x+dx)^3-10(x+dx)^2-12

Answer 79

yeah I got up to 15x^2+15xdx+25dx^2-20x+10dx, but idk if i did it right, or made mistakes along the way. or what to do next.

Answer 80

no...there must be some cubic term in ur expression after simplification

Answer 81

let's do the calculation together...there are two ways

Answer 82

ok

Answer 83

5(x+dx)^3-10(x+dx)^2-12 = 5(x+dx)^2 *{ (x+dx)-2} -12

Answer 84

i have taken 5(x+dx)^2 common from 1st two terms...

Answer 85

it's ok?

Answer 86

seems reasonable...

Answer 87

x+dx-1?

Answer 88

how did u get it?

Answer 89

i mean 2

Answer 90

yes

Answer 91

ok that makes it a lot simpler

Answer 92

5(x+dx)^2 *{ (x+dx)-2} -12
=5( x^2 + dx^2 +2 x dx)*( x+dx -2) -12

Answer 93

now using distributive law do the multiplication..

Answer 94

5 x^2 +5 dx^2 +10 x dx)*( x+dx -2) -12

Answer 95

is the -12 diistributed as well?

Answer 96

5( x^2 + dx^2 +2 x dx)*( x+dx -2) -12
5{ x^2*( x+dx -2) + dx^2* ( x+dx -2) + 2 x dx*( x+dx -2) } -12

Answer 97

oh you distribute the 5 through all that?

Answer 98

no...i didnt distribute 5 yet...it's still outside..