# using limit derivative?

• December 10th 2008, 08:59 PM
mathseek
using limit derivative?
how to solve using limit definition of the derivative for f(x)=2x^3+5x^2-7x-4

my ans:

6x^2+10x-7

plse correct me...
• December 10th 2008, 09:18 PM
Prove It
Quote:

Originally Posted by mathseek
how to solve using limit definition of the derivative for f(x)=2x^3+5x^2-7x-4

my ans:

6x^2+10x-7

plse correct me...

It's correct...
• December 10th 2008, 09:23 PM
mathseek
limit def
than what is limit definition??
• December 10th 2008, 09:27 PM
o_O
Definition of a derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
• December 10th 2008, 09:32 PM
mathseek
limit def
so when i answer this question above....i just give my ans will do right?
• December 10th 2008, 09:40 PM
Prove It
Quote:

Originally Posted by mathseek
so when i answer this question above....i just give my ans will do right?

No - I thought you had gone through the rigmarole of using the limit definition and gotten that answer.

$f(x) = 2x^3 + 5x^2 -7x - 4$

$f(x + h) = 2(x + h)^3 + 5(x + h)^2 - 7(x + h) - 4$

$= 2(x^3 + 3x^2h + 3xh^2 + h^3) + 5(x^2 + 2xh + h^2) - 7x -7h -4$

$= 2x^3 + 6x^2h + 6xh^2 + 2h^3 + 5x^2 + 10xh + 5h^2 - 7x - 7h - 4$

So $f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 + 10xh + 5h^2 - 7h$

$= h[6x^2 + 6xh + 2h^2 + 10x + 5h - 7]$.

$\frac{f(x+h) - f(x)}{h} = 6x^2 + 6xh + 2h^2 + 10x + 5h - 7$

$\lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = 6x^2 + 6x(0) + 2(0)^2 + 10x + 5(0) - 7$

$= 6x^2 + 10x - 7$.
• December 10th 2008, 09:42 PM
o_O
Well you have to show your work probably and you do so by going through the grunt work of expanding and simplifying in order to evaluate the limit:

\begin{aligned}f'(x) & = \lim_{h \to 0} \frac{2(x+h)^3+5(x+h)^2-7(x+h) - 4 \ - \ \left(2x^3 + 5x^2 - 7x - 4\right)}{h} \\ & = \lim_{h \to 0} \frac{2(x^3 + 2x^2h + 2xh^2 + h^3) +5(x^2 + 2xh + h^2) - \cdots}{h} \end{aligned}

Eventually you can factor out an h from the top to cancel it with the bottom and you should get your answer. Like I said, a lot of grunt work.