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bigcat
Group Title
find the derivative of f(x)=x^2 + x 3 using the definition of the derivative
 one year ago
 one year ago
bigcat Group Title
find the derivative of f(x)=x^2 + x 3 using the definition of the derivative
 one year ago
 one year ago

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Callisto Group TitleBest ResponseYou've already chosen the best response.3
\[f'(x)\]\[=\lim_{h \rightarrow 0}\frac{f(x+h)  f(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{[(x+h)^2 + (x+h) 3] (x^2 + x 3)}{h}\]Can you simplify the fraction first?
 one year ago

bigcat Group TitleBest ResponseYou've already chosen the best response.0
I used x^2 + 2(x)(h) + h^2 + x +h 3 x^2 x +3 = 2x + x ??
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
x^2 + 2(x)(h) + h^2 + x +h 3 x^2 x +3 < correct for denominator. But it's not equal to 2x + x... \[x^2 + 2(x)(h) + h^2 + x +h 3 x^2 x +3\]Group the like terms together: \[x^2 x^2 + x x 3 +3+ h^2+ 2(x)(h)+h\] Now, can you simplify the above expression
 one year ago

bigcat Group TitleBest ResponseYou've already chosen the best response.0
basically factor out the h and H(h+2x+1) divided by the h on denominator and = h + 2x + 1?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
And you got: \[\lim_{h \rightarrow 0} (h+2x+1)\]Can you evaluate the limit?
 one year ago

bigcat Group TitleBest ResponseYou've already chosen the best response.0
Would be 2x + 1 ??? Because h is going to 0
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
Indeed it is!
 one year ago

bigcat Group TitleBest ResponseYou've already chosen the best response.0
Holy Cow! It came together finally!! Thank you very much for your help! I would have been stuck!!!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
You're welcome :)
 one year ago

bigcat Group TitleBest ResponseYou've already chosen the best response.0
Have a great day!!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.3
You too! :)
 one year ago
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