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anonymous
 one year ago
I have no idea how to solve this. Can someone please help me for a medal? :)
(posting below)
anonymous
 one year ago
I have no idea how to solve this. Can someone please help me for a medal? :) (posting below)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ f(x+Δx)−f(x) }{ Δx }\] if f(x) = x^2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1compute as \(\huge \frac{(x + \Delta x)^2  x^2}{\Delta x}\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so expand out the \(\large (x + \Delta x)^2\) in the numerator :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know how to get to the expanding part, but I am not sure how to expand the numerator. Could you show me how?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it end up looking like this... \[\frac{ x^{2} + Δx^{2}x^{2} }{ Δx } \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1first expand out \( (x+Δx)^2\) on its own that's \( (x+Δx) \times (x+Δx)\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1440705634106:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That makes sense, but how do you multiple an "Δx" by an "x"

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(x \times \Delta x = x \ \Delta x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the numerator would end up being... \[\frac{ x^{2}+2xΔx+Δx^{2}x^{2} }{ Δx }\] which would simplify to... \[\frac{ 2xΔx+Δx^{2} }{ Δx }\] is that correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ Δx(2x+Δx) }{ Δx }\] And the final answer would be... 2x+Δx ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yay! well done learning calculus? happy days!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much. :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1any time! thank you.
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