Derivatives

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kidmo87

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Sep 10, 2012
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Hey everyone. Obviously my question is about derivatives. On my homeworks, when finding the derivatives, it usually starts with x^2 or x^3. But my teacher gave me a problem that starts with a number, then x^3. The problem is 2x^3-4x^2+3. Ive been trying to solve this problem, but get stuck when its time to cancel out. Honestly i think i made errors from the beginning. What i have is:
f'(x)=lim [2(x+h)^3 -4(x+h)^2 +3]-[2x^3-4x^2+3]
 h(arrow)0 h

f'(x)=lim [2(x^3+2xh+h^3-4x^2+h^2+3]-[2x^3-4x^2+3]
 h(arrow)0 h From this point, i feel that something is wrong.


I would greatly appreciate is someone can help me out. My guess is that from the problem, im not plugging in 2x^3 or 4x^2 correctly.
 
kidmo87 said:
Hey everyone. Obviously my question is about derivatives. On my homeworks, when finding the derivatives, it usually starts with x^2 or x^3. But my teacher gave me a problem that starts with a number, then x^3. The problem is 2x^3-4x^2+3. Ive been trying to solve this problem, but get stuck when its time to cancel out. Honestly i think i made errors from the beginning. What i have is:
f'(x)=lim [2(x+h)^3 -4(x+h)^2 +3]-[2x^3-4x^2+3]
h(arrow)0 h

f'(x)=lim [2(x^3+2xh+h^3-4x^2+h^2+3]-[2x^3-4x^2+3]
h(arrow)0 h From this point, i feel that something is wrong.


I would greatly appreciate is someone can help me out. My guess is that from the problem, im not plugging in 2x^3 or 4x^2 correctly.
Click to expand...


You expanded \(\displaystyle (x+h)^3\) and \(\displaystyle (x+h)^2\) incorrectly.

\(\displaystyle (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) and \(\displaystyle (a + b)^2 = a^2 + 2ab + b^2\)

Apply these formulas accordingly and then see what you get.
 
Need to start with expanding (x+h)^3 correctly. If you can do that you can probably finish the rest.
I'll start, (x+h)^3 = (x+h)(x+h)(x+h) = (x^2 + 2xh + h^2)(x+h) = ?
Continue...
 
Thanks. So did i do the first line right, or do i need to work on that too?
 
hey guys, so ive been trying to solve this problem, and still coming across some issues. This is where im at soo far.

2(x^3+3x^2h+3xh^2+h^3)-4(x^2+2xh+h^2)+3 -(2x^3-4x^2+3)

= 2x^3+6x^2h+6xh^2+2h^2 -4x^2-8xh-4h^2 +3 -2x^3+4x^2 -3

After crossing out, im left with:

6x^2h+6xh^2+2h^3-8xh-4h^2
 h

Could you guys help me with where i went wrong with this, thanks.
 
kidmo87 said:
hey guys, so ive been trying to solve this problem, and still coming across some issues. This is where im at soo far.

2(x^3+3x^2h+3xh^2+h^3)-4(x^2+2xh+h^2)+3 -(2x^3-4x^2+3)

= 2x^3+6x^2h+6xh^2+2h^2 -4x^2-8xh-4h^2 +3 -2x^3+4x^2 -3

After crossing out, im left with:

6x^2h+6xh^2+2h^3-8xh-4h^2
 h

Could you guys help me with where i went wrong with this, thanks.
Click to expand...
Why do you think you went wrong? Just work out the algebra.

\(\displaystyle f(x) = 2x^3 - 4x^2 + 3 \implies\)

\(\displaystyle f(x + h) = 2(x + h)^3 - 4(x + h)^2 + 3 = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 4(x^2 + 2xh + h^2) + 3 =\)

\(\displaystyle 2x^3 - 4x^2 + 3 + 6x^2h + 6xh^2 + 2h^3 - 8xh - 4h^2 \implies\)

\(\displaystyle f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 8xh - 4h^2 = h(6x^2 - 8x) + h^2(6x + 2h - 4) \implies \)

\(\displaystyle \dfrac{f(x + h) - f(x)}{h} = 6x^2 - 8x + h(6x + 2h - 4).\)

Now take the limit.
 
ok, i guess i was on the right track. so when i take the limit, do i have to plug in 0 for h since h(arrow)o ?
approaches
 
kidmo87 said:
ok, i guess i was on the right track. so when i take the limit, do i have to plug in 0 for h since h(arrow)o ?
approaches
Click to expand...
There are many different ways to find a limit but the simplest is if f(x) is continuous at x= a, then \(\displaystyle \lim_{h\to a} f(x)= f(a)\)
 
Alright, thank you soo much guys. I know its probably been a hastle, and a little frustrating trying to help me, but its greatly appreciated. Thanks.
 
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