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Lol, derivative?

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Other answers:

yes i think so that doesnt cancel out tho -.- im making a mistake
\[\huge \lim_{h \rightarrow 0}\frac{[(x+h)^{3}-3]-(x^{3}-3)}{h}\] is this it?
yupp ! how do you make it big like that btw ?
uhh, before typing in your equations and stuff, put in \huge
thanks:D
This looks like you're taking the derivative of \[\huge x^{3} - 3\]
yup that's right !
And you're to get the derivative using limits? lol, that's harsh... :D
yess i have to
You'll realise soon that there are faster ways of getting the derivative... anyway, just evaluate the numerator and show me what you get...
x^3 's cancel 3's cancel... what's left?
3x^2h + 3xh^2 +h^3 all divided by h factor out an 'h' in the numerator cancel with the 'h' in the denom. now you can take the limit without issues.
\[\huge \ \frac{ x^3+3x^2h+3xh^2-3-x^2+3 }{h }\]
no.
i have an x^3 and x^2 ? :S
no.
Well that changes everything...
\[\frac{ f(x+h) - f(x) }{h }\]
what's f(x)?
x^3 -3
\[\frac{(x+h)^3 -3 - ( x^3 -3) }{h }\]
no f(x)=x^2-3
thats why it doesnt cancel out
you expanded correctly: \[\frac{ x^3 +3x^2h +3xh^2 +h^3 -3 - (x^3 -3) }{h }\]
ok
then why are you cubing (x+h) ?
ohhh i should square it *facepalm*
\[\frac{ (x+h)^2 -3 -(x^2-3) }{h }\]
can you help me with this one: find the equation of the tangent \[y=\frac{ x-2 }{ x+2 }\] when x=0
\[\huge \lim_{h \rightarrow 0} \frac{ (x+h)-2 }{ (x+h)+2 } -\frac{ x-2 }{ x+2 }\]
Equation of the tangent involves derivatives, right?
yes
Quotient rule? Ever dabbled with it?
no we're not allowed to use that yet -.-
Here, I don't think I have the attention span to type this out o.O http://www.math.hmc.edu/calculus/tutorials/quotient_rule/proof.pdf
You're allowed to use anything as long as you can prove it ;)
my tests always state "Only using method thought in class" i know that quotient rule is so much faster
and we we're never thought that in class yet
Well, don't despair :D I'll see what I can do... To get the equation of the tangent line, first we need a slope. to get that slope, we need the derivative. So what's the derivative (go ahead and express it first in limit-form)
\[\huge \lim_{h \rightarrow o}\frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 }\]
nope... you forgot that all that must also be divided by h... small details, but no less important...
ohh i did :$
tsk \[\huge \lim_{h \rightarrow 0}\frac{\frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 }}{h}\]
ok yupp
Right, well this can also be written as \[\lim_{h \rightarrow 0}\frac{1}{h}\left( \frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 } \right)\]
oh okayy
now \[\lim_{h \rightarrow 0}\frac{1}{h}\frac{\left[ \left( x+h \right)-2 \right]\left( x+2 \right)-\left[ \left( x+h \right)+2 \right]\left( x-2 \right)}{\left[ \left( x+h \right)+2 \right]\left( x+2 \right)}\] o.O
wait, how did you get rid of those fractions ?
Like magic :P \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]
Get it now?
yeaaah ! i do
and then cancelling
Hey, there's nothing you can cancel...
then expansion and cancelling out the like terms
Nothing that would you do you much good, anyway... Here's a tricky bit of manipulation \[\lim_{h \rightarrow 0}\frac{1}{h}\frac{\left[ \left( x+h \right)-2 \right]\left( x+2 \right)-(x-2)(x+2)+(x-2)(x+2)-\left[ \left( x+h \right)+2 \right]\left( x-2 \right)}{\left[ \left( x+h \right)+2 \right]\left( x+2 \right)}\]
wow o.O
All right, scratch everything, I overlooked one important detail...
You only need to get its slope on the specific case where x = 0 right?
Use this definition of derivative instead: \[f'(c)=\lim_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}\]
whats c ?
Sorry for this extremely late reply, I was busy with something :( c is 0.

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