kk then, what is deriative??

- anonymous

kk then, what is deriative??

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

study of how a function change, or tangent line

- amistre64

derivative is what is used to describe how one thing changes with respect to another

- anonymous

hmmm.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

It is like a slope

- amistre64

in the equation: y = 3x+2, the rate at which y changes with respect to x is '3'

- amistre64

for every change in 'x', 'y' changes by 3

- anonymous

hmm then u mean 2y=3x+3+2 ??

- amistre64

not quite

- anonymous

or 2y=6x+2?? (the 2 is const in this equetion?

- amistre64

if you multiply the left by 2 the rights gotta be multiplied by 2 as well

- amistre64

thats more of a scaling notion

- amistre64

the derivative has to do with the change in y with respect to x when there is 'no' change in x

- anonymous

hmmm

- amistre64

\[\lim_{h -> 0}\frac{(f(x+h)-f(x)}{(x+h)-(x)}\]

- amistre64

the derivative of: 3x+2 is then
(3(x+h)+2) - (3x+2)
----------------- ; when h=0
x+h-x

- amistre64

3x+3h+2 -3x-2
-------------- ; when h=0
h
3h
--- ; when h=0
h
3 ; when h=0

- anonymous

hmmm

- amistre64

so the rate of change of a linear function is just the slope of the line

- anonymous

hmmm

- amistre64

the rate of change at any given point of a quadratic function is the slope of the line at a given point that is defined as the lim{h ->0} (f(x+h)-f(x)//h)

- amistre64

x^2+x-3 ; how fast is y changing when x = 4?

- anonymous

can you answer it? i think i can'T do it yet

- amistre64

\[\lim_{h -> 0}\frac{ [(x+h)^2 +(x+h)-3]-[x^2+x-3]}{x+h-x}\]

- amistre64

\[\lim{x^2+h^2 +2xh +x+h-3-x^2-x-3 \over h}\]

- anonymous

okay give me a min

- anonymous

getting my pencil and notebook

- amistre64

that last 3 shoud be +3

- anonymous

kk

- anonymous

h^2 + 2xh + h / h

- amistre64

\[(x^2-x^2)+(x-x)+(-3+3)+(2xh+h) \over h\]

- amistre64

yes, i missed the ^2 on the h; you are roght

- anonymous

:D

- amistre64

\[\lim_{h \rightarrow 0}\left({h^2\over h}+{2xh\over h}\right)\]

- anonymous

now we give x 3??

- amistre64

when x=4 :)

- amistre64

and I think we dropped an h someplace in the writing

- anonymous

ahh missed 4 xd

- anonymous

yes why don'T we do it one a more simple func??

- amistre64

\[\lim_{h ->0}({2xh\over h}+{h\over h}+{h^2\over h})\]

- amistre64

the simple function was the linear lol

- amistre64

now when h=0 we get the equation of the derivative:
2x+1

- amistre64

at x=4; y is changing at a rate of 9 with respect to each x

- amistre64

the quick way after being taught all the limit definitions and the long way; is to use the power rule:
x^2 +x -3 -> 2x +1

- anonymous

hmm ^^ so that can we use this for another thing?

- anonymous

with another func i mean

- anonymous

(don't got the power rule can you explain?)

- amistre64

yes; the derivative of position tells you velocity at a given point; the derivative of velocity tells you the acceleration at any given point

- amistre64

the power rule is just the 'pattern' that occurs from doing derivatives the long way..

- anonymous

oO

- amistre64

\[Cx^n \implies C*nx^{n-1} \]
\[5x^3 \implies 5*3x^{3-1} \iff 15x^2\]

- amistre64

which way would you do derivatives? the limit way or the power rule?

- amistre64

\[\begin{array}c 5x^4&+x^2&-4x^0\\20x^3&+2x&+0\end{array}\]

- anonymous

hmm
i think limit way

- amistre64

ok; do that the limit way...
\[\frac{[5(x+k)^4+(x+h)^2-4]-[5x^4+x^2-4]}{h}\]

- anonymous

which equetion we doing??

- amistre64

f(x) = \(5x^4 +x^2 -4\) the limit way

- amistre64

except for it should say 5(x+h)^4 lol

- amistre64

the power rule is quicker, easier, and just as precise

- anonymous

ok :=)

- anonymous

then do it in power rule ^^ if its that trustworthy

- amistre64

the power rule is the 'patter' that you discover after doing about a thousand limit ways

- anonymous

it was Cx^n = C x nx^n-1

- amistre64

yes

- anonymous

ok

- anonymous

then for 5(x+4)^4 its >>>> 5 x 4(x)^3 ???

- anonymous

hey you there?

- amistre64

ok.... you trying to use the x+h in that?

- anonymous

no i tried to use power rule that i did C.nx^n-1

- amistre64

5x^4 ; try it on that one ...

- anonymous

oka wait

- anonymous

5 x 4(x)^3

- anonymous

5 . 4.x^3

- anonymous

true?

- amistre64

true; now put it all together; 20x^3

- anonymous

yess i got it??

- amistre64

you did; the power rule takes a complicated form and reduces it to simple math

- anonymous

yes its really usefull :) to you, am i good at maths?

- amistre64

you can be good at maths if you wanna be :)

- anonymous

:) really want to be ^^ i want to learn highschool units before i go to hichschool ^^

- anonymous

really thanks amistre :)

- amistre64

youre welcome :) keep up the practics

- anonymous

normally just searching questions on Hz. google xd

- amistre64

some good free practice can be had at interactmath.com

- anonymous

ok ty

Looking for something else?

Not the answer you are looking for? Search for more explanations.