Another, use definition as a derivative (as a limit) problem 2

  • #1
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use definition as a derivative (as a limit) to show that

Dx (x(x+1)) = 2x + 1, x [itex]\in[/itex] ℝ

Would I need to find the derivative first?

f'(x) = 2
 

Answers and Replies

  • #2
vela
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No, you're supposed to calculate the derivative of x(x+1) using the definition of the derivative.
 
  • #3
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So the derivative of x(x+1) is just 1?
 
  • #4
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No, it is not. Since you are supposed to show (by using the limit definition of the derivative) that the derivative of x(x + 1) = 2x + 1, why would you think that the derivative is 1?
 
  • #5
HallsofIvy
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Do you know what the definition of derivative (as a limit) is?
 
  • #6
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Is this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?
 
  • #7
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No, it is not. Since you are supposed to show (by using the limit definition of the derivative) that the derivative of x(x + 1) = 2x + 1, why would you think that the derivative is 1?
Whoops I meant 2x+1 = 2x^1-1 + 0 = 2
 
  • #8
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Is this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?
This would give you f'(a). For f'(x), the definition is this limit:
[tex]\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}[/tex]
 
  • #9
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Whoops I meant 2x+1 = 2x^1-1 + 0 = 2
2x + 1 ≠ 2x1 - 1.

Note that what you wrote in the middle expression above is interpreted as 2x1 - 1 + 0 = 2x - 1, and that's probably not what you intended. If you have an exponent that is an arithmetic expression, USE PARENTHESES.

Also, in what you wrote, you are apparently taking a derivative, but your notation gives no hint that you're doing this. What you should have written is
d/dx(2x + 1) = 2 + 0 = 2.
 
  • #10
vela
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Is this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?
The denominator should be h, not f(h). Mark provided you with the correct definition. So you want to evaluate
[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{[(x+h)((x+h)+1)] - x(x+1)}{h}[/tex]Multiply out the top, cancel terms, simplify, and then take the limit. If you do it correctly, you'll find you end up with 2x+1.
 
  • #11
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I dont know how but I ended getting (2x-x+1) using the derivative (as a limit) formula.

[itex]\stackrel{Lim}{h→0}[/itex] [itex]\frac{(x^2+2xh+h^2+1)-x(x+1)}{h}[/itex]

[itex]\stackrel{Lim}{h→0}[/itex] (2x-x+1)
 
Last edited:
  • #12
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I dont know how but I ended getting (2x-x+1) using the derivative (as a limit) formula.

[itex]\stackrel{Lim}{h→0}[/itex] [itex]\frac{(x^2+2xh+h^2+1)-x(x+1)}{h}[/itex]
What you're starting with is incorrect. It should be
[tex]\lim_{h \to 0}\frac{(x + h)(x + h + 1) - x(x + 1)}{h}[/tex]

It would make life easier if you rewrote x(x + 1) as x2 + x.
[itex]\stackrel{Lim}{h→0}[/itex] (2x-x+1)
 
  • #13
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Ahhh I ended getting 2x+1, thank you.
 

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