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Ppl of the world.. here comes the Next question -..-

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its about the definition of the derivative >>>
were all beaming with anticipation...
Apply the definition of the d derivative to compute \[f \prime (x) \] where \[f(x)=3x ^{2}+x-2\]

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Other answers:

i know .. \[f'(x)= 6x +1\] .. but my teacher said do it with details like long way to get this answer
-..- i don't know how
i see well |dw:1349804778192:dw|
XD ur hand-writting is Funnie XP
lim delta x tends to 0
O_O sorry didn't mean to .. oops im sorry >> >_< sowwie
you have a function of x F(X)=3x²+x−2 to differentiate you multiply the coefficient of the unknown by the power of the unknown and decrease the power of the unknown by 1 for example f(x)=x² f'(x)=2(1)x^2-1=2x so f(x)=x f'(x)=1(1)x^1-1=x^0=1 if you have f(x)=2x² f'(x)=2(2)x^2-1=4x does this help? |dw:1349804820461:dw|
np @Alice_Dump did u get that ?
-..- ppl brb .. gemme 5 minutes i need to do something .
wait okie >_<
\[f'(x)=2\times 3x^{2-1} +1\times x^{1-1}-0\times2x^{0-1}=6x +1\]
\[f(x)=3x^2+x-2\] \[f(x+h)=3(x+h)^2+(x+h)-2=3(x^2+2xh+h^2)+x+h-2\] \[=3x^2+6xh+3h^2+x+h-2\] \[f(x+h)-f(x)=3x^2+6xh+3h^2+x+h-2-(32x^2+x-2)\] \[=6xh+h+3h^2\]
2 being constant will disappear
divide by \(h\) and get \[\frac{f(x+h)-f(x)}{h}=6x+1+3h\] take the limit as \(h\to 0\) get \(6x+1\)
you needed it done by hand right? not power rule it is a bunch of algebra, that is all only last step is calc, and that is taking the limit by replacing \(h\) by \(0\)
indeed @satellite73 has illustrated the given problem nicely with the formula i have given you ...i think that clears evrything
question said "Apply the definition of the derivative to compute"
@AravindG thanks
yw :)
:} back ..
i got the question now :] thnx ppl ^_^'

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