Can someone verify my answer for the problem in my post below

  • Thread starter kendal12
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kendal12

Main Question or Discussion Point

ok, so I worked out the problem and got 2x+a... is this correct? For some reason I dont think it is, but I thought I worked it out right... I just want to know if I'm on the right track so I dont do all 50 problems wrong
 

Answers and Replies

1,233
1
which problem?
 
kendal12
sorry, i guess I posted a different problem below...

lim ( (x+∆x)^2 - x^2 ) / x
∆x approaches 0

Sorry about that
 
CRGreathouse
Science Advisor
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kendal12 said:
lim ( (x+∆x)^2 - x^2 ) / x
∆x approaches 0
Do you mean
lim ( (x+∆x)^2 - x^2 ) / ∆x
∆x approaches 0?

If so, you get 2x. Otherwise, if you meant the problem as you typed it, you get 0 (or undefined if x=0).
 
80
0
Your problem is a verification of the definition of derivative.
In particular [tex]\frac{d}{dx}\left[x^2 \right] = \lim _{\Delta x\rightarrow 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x} =
\lim _{\Delta x\rightarrow 0} \frac{x^2 + 2x\Delta x + (\Delta x)^2 - x^2}{\Delta x} = \lim _{\Delta x\rightarrow 0} \frac{2x\Delta x + (\Delta x)^2 }{\Delta x} = \lim _{\Delta x\rightarrow 0} (2x+ \Delta x) = 2x [/tex]

Notice that in general (for other functions) both right and left limit have to exist in order to say that the derivative exists. In this case you have a simple polinomyal problem and this is not an issue.

If you do not like the notation [tex] \Delta x[/tex] for a small number, you can use the equivalent notation:

[tex]\frac{d}{dx}\left[x^2 \right] = \lim _{\varepsilon \rightarrow 0} \frac{(x+\varepsilon)^2 - x^2}{\varepsilon} =
\lim _{\varepsilon\rightarrow 0} \frac{x^2 + 2x\varepsilon + (\varepsilon)^2 - x^2}{\varepsilon} = \lim _{\varepsilon\rightarrow 0} \frac{2x\varepsilon + (\varepsilon)^2 }{\varepsilon} = \lim _{\varepsilon\rightarrow 0} (2x+ \varepsilon) = 2x [/tex]
 
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