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

[CrossFit415]

use definition as a derivative (as a limit) to show that

D_x (x(x+1)) = 2x + 1, x \in ℝ

Would I need to find the derivative first?

f'(x) = 2

 

[vela]

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

 

[CrossFit415]

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

 

[Mark44]

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?

 

[HallsofIvy]

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

 

[CrossFit415]

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

 

[CrossFit415]

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

 

[Mark44]

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:
\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

 

[Mark44]

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

2x + 1 ≠ 2x^{1 - 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.

 

[vela]

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
\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}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.

 

[CrossFit415]

I don't know how but I ended getting (2x-x+1) using the derivative (as a limit) formula.

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

\stackrel{Lim}{h→0} (2x-x+1)

 

[Mark44]

I don't know how but I ended getting (2x-x+1) using the derivative (as a limit) formula.

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

What you're starting with is incorrect. It should be
\lim_{h \to 0}\frac{(x + h)(x + h + 1) - x(x + 1)}{h}

It would make life easier if you rewrote x(x + 1) as x² + x.

\stackrel{Lim}{h→0} (2x-x+1)

 

[CrossFit415]

Ahhh I ended getting 2x+1, thank you.