- #1

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D

_{x}(x(x+1)) = 2x + 1, x [itex]\in[/itex] ℝ

Would I need to find the derivative first?

f'(x) = 2

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- Thread starter CrossFit415
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- #1

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D

Would I need to find the derivative first?

f'(x) = 2

- #2

vela

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No, you're supposed to calculate the derivative of x(x+1) using the definition of the derivative.

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So the derivative of x(x+1) is just 1?

- #4

Mark44

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- #5

HallsofIvy

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Do you know what the definition of derivative (as a limit) **is**?

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Is this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?

- #7

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Whoops I meant 2x+1 = 2x^1-1 + 0 = 2

- #8

Mark44

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This would give you f'(a). For f'(x), the definition is this limit:Is this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?

[tex]\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}[/tex]

- #9

Mark44

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2x + 1 ≠ 2xWhoops I meant 2x+1 = 2x^1-1 + 0 = 2

Note that what you wrote in the middle expression above is interpreted as 2x

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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The denominator should be h, not f(h). Mark provided you with the correct definition. So you want to evaluateIs this the definition, Lim h -> 0 [f(a+h) - f(a)] / [f(h)] ?

[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)

[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)

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- #12

Mark44

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What you're starting with is incorrect. It should beI 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]

[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 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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