anonymous
  • anonymous
kk then, what is deriative??
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
study of how a function change, or tangent line
amistre64
  • amistre64
derivative is what is used to describe how one thing changes with respect to another
anonymous
  • anonymous
hmmm.

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anonymous
  • anonymous
It is like a slope
amistre64
  • amistre64
in the equation: y = 3x+2, the rate at which y changes with respect to x is '3'
amistre64
  • amistre64
for every change in 'x', 'y' changes by 3
anonymous
  • anonymous
hmm then u mean 2y=3x+3+2 ??
amistre64
  • amistre64
not quite
anonymous
  • anonymous
or 2y=6x+2?? (the 2 is const in this equetion?
amistre64
  • amistre64
if you multiply the left by 2 the rights gotta be multiplied by 2 as well
amistre64
  • amistre64
thats more of a scaling notion
amistre64
  • amistre64
the derivative has to do with the change in y with respect to x when there is 'no' change in x
anonymous
  • anonymous
hmmm
amistre64
  • amistre64
\[\lim_{h -> 0}\frac{(f(x+h)-f(x)}{(x+h)-(x)}\]
amistre64
  • amistre64
the derivative of: 3x+2 is then (3(x+h)+2) - (3x+2) ----------------- ; when h=0 x+h-x
amistre64
  • amistre64
3x+3h+2 -3x-2 -------------- ; when h=0 h 3h --- ; when h=0 h 3 ; when h=0
anonymous
  • anonymous
hmmm
amistre64
  • amistre64
so the rate of change of a linear function is just the slope of the line
anonymous
  • anonymous
hmmm
amistre64
  • 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
  • amistre64
x^2+x-3 ; how fast is y changing when x = 4?
anonymous
  • anonymous
can you answer it? i think i can'T do it yet
amistre64
  • amistre64
\[\lim_{h -> 0}\frac{ [(x+h)^2 +(x+h)-3]-[x^2+x-3]}{x+h-x}\]
amistre64
  • amistre64
\[\lim{x^2+h^2 +2xh +x+h-3-x^2-x-3 \over h}\]
anonymous
  • anonymous
okay give me a min
anonymous
  • anonymous
getting my pencil and notebook
amistre64
  • amistre64
that last 3 shoud be +3
anonymous
  • anonymous
kk
anonymous
  • anonymous
h^2 + 2xh + h / h
amistre64
  • amistre64
\[(x^2-x^2)+(x-x)+(-3+3)+(2xh+h) \over h\]
amistre64
  • amistre64
yes, i missed the ^2 on the h; you are roght
anonymous
  • anonymous
:D
amistre64
  • amistre64
\[\lim_{h \rightarrow 0}\left({h^2\over h}+{2xh\over h}\right)\]
anonymous
  • anonymous
now we give x 3??
amistre64
  • amistre64
when x=4 :)
amistre64
  • amistre64
and I think we dropped an h someplace in the writing
anonymous
  • anonymous
ahh missed 4 xd
anonymous
  • anonymous
yes why don'T we do it one a more simple func??
amistre64
  • amistre64
\[\lim_{h ->0}({2xh\over h}+{h\over h}+{h^2\over h})\]
amistre64
  • amistre64
the simple function was the linear lol
amistre64
  • amistre64
now when h=0 we get the equation of the derivative: 2x+1
amistre64
  • amistre64
at x=4; y is changing at a rate of 9 with respect to each x
amistre64
  • 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
  • anonymous
hmm ^^ so that can we use this for another thing?
anonymous
  • anonymous
with another func i mean
anonymous
  • anonymous
(don't got the power rule can you explain?)
amistre64
  • 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
  • amistre64
the power rule is just the 'pattern' that occurs from doing derivatives the long way..
anonymous
  • anonymous
oO
amistre64
  • amistre64
\[Cx^n \implies C*nx^{n-1} \] \[5x^3 \implies 5*3x^{3-1} \iff 15x^2\]
amistre64
  • amistre64
which way would you do derivatives? the limit way or the power rule?
amistre64
  • amistre64
\[\begin{array}c 5x^4&+x^2&-4x^0\\20x^3&+2x&+0\end{array}\]
anonymous
  • anonymous
hmm i think limit way
amistre64
  • amistre64
ok; do that the limit way... \[\frac{[5(x+k)^4+(x+h)^2-4]-[5x^4+x^2-4]}{h}\]
anonymous
  • anonymous
which equetion we doing??
amistre64
  • amistre64
f(x) = \(5x^4 +x^2 -4\) the limit way
amistre64
  • amistre64
except for it should say 5(x+h)^4 lol
amistre64
  • amistre64
the power rule is quicker, easier, and just as precise
anonymous
  • anonymous
ok :=)
anonymous
  • anonymous
then do it in power rule ^^ if its that trustworthy
amistre64
  • amistre64
the power rule is the 'patter' that you discover after doing about a thousand limit ways
anonymous
  • anonymous
it was Cx^n = C x nx^n-1
amistre64
  • amistre64
yes
anonymous
  • anonymous
ok
anonymous
  • anonymous
then for 5(x+4)^4 its >>>> 5 x 4(x)^3 ???
anonymous
  • anonymous
hey you there?
amistre64
  • amistre64
ok.... you trying to use the x+h in that?
anonymous
  • anonymous
no i tried to use power rule that i did C.nx^n-1
amistre64
  • amistre64
5x^4 ; try it on that one ...
anonymous
  • anonymous
oka wait
anonymous
  • anonymous
5 x 4(x)^3
anonymous
  • anonymous
5 . 4.x^3
anonymous
  • anonymous
true?
amistre64
  • amistre64
true; now put it all together; 20x^3
anonymous
  • anonymous
yess i got it??
amistre64
  • amistre64
you did; the power rule takes a complicated form and reduces it to simple math
anonymous
  • anonymous
yes its really usefull :) to you, am i good at maths?
amistre64
  • amistre64
you can be good at maths if you wanna be :)
anonymous
  • anonymous
:) really want to be ^^ i want to learn highschool units before i go to hichschool ^^
anonymous
  • anonymous
really thanks amistre :)
amistre64
  • amistre64
youre welcome :) keep up the practics
anonymous
  • anonymous
normally just searching questions on Hz. google xd
amistre64
  • amistre64
some good free practice can be had at interactmath.com
anonymous
  • anonymous
ok ty

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