# Could use some Derivation help, please

• Apr 12th 2008, 05:44 AM
Hanga
Could use some Derivation help, please
Okay so I just started with derivation.. it's pretty damn fun but the Swedish standard math book sucks completly because they can't give a decent example even if their life depended on it.
Anywho here is problem.

Calculate f(4) by using numeric derivation, if;

a) f(x) = 3x^4 -350x + 129

b) f(x) = x / (x^2 + 1)

c) f(x) = 6.5 * 10^(0,0086x)

On a) I tried to use the basic lim h > 0, it did work but there was so many damn numbers It's like made for you to screw up.
f(4+h) - f(4)
3(4+h)^4 - 350(4+h) + 129

You see it right? (4+h)^4 is pure pain to keep track of.. is there an easier way for me to simplify this so the calculation won't get so long?

On b) I got stuck completly.
f(4+h) -f(4) = ((4+h) / ((4+h)^2 +1))) - 4 / (4^2 + 1)

Could someone help me expand this and give me some derivation tips?

P.S Thx to your help I got a B+ on my last math thest which I've never had before! Thanks ALOT!
D.S
• Apr 12th 2008, 05:51 AM
Moo
Hello,

D'you know the general formulas for derivations ?

The derivative of a sum is the sum of the derivatives etc...

Or you can do it simply :

$\displaystyle f(4+h)-f(4)=3(4+h)^4-350(4+h)+129-3(4)^4+350(4)-129$$\displaystyle =3\left((4+h)^4-4^4 \right)-350(4+h-4)=3\left((4+h)^4-4^4 \right)-350h Let's look what is \displaystyle \left((4+h)^4-4^4 \right) \displaystyle (4+h)^4=((4+h)^2)^2 \displaystyle 4^4=(4^2)^2=16^2 So \displaystyle \left((4+h)^4-4^4 \right)=a^2-b^2 with \displaystyle a=(4+h)^2=16+8h+h^2 and \displaystyle b=16 \displaystyle \left((4+h)^4-4^4 \right) =a^2-b^2=(a-b)(a+b) =$$\displaystyle (16+8h+h^2-16)(16+8h+h^2+16)=(8h+h^2)(32+8h+h^2)$
• Apr 12th 2008, 06:04 AM
Mathstud28
Ok
Quote:

Originally Posted by Hanga
Okay so I just started with derivation.. it's pretty damn fun but the Swedish standard math book sucks completly because they can't give a decent example even if their life depended on it.
Anywho here is problem.

Calculate f(4) by using numeric derivation, if;

a) f(x) = 3x^4 -350x + 129

b) f(x) = x / (x^2 + 1)

c) f(x) = 6.5 * 10^(0,0086x)

On a) I tried to use the basic lim h > 0, it did work but there was so many damn numbers It's like made for you to screw up.
f(4+h) - f(4)
3(4+h)^4 - 350(4+h) + 129

You see it right? (4+h)^4 is pure pain to keep track of.. is there an easier way for me to simplify this so the calculation won't get so long?

On b) I got stuck completly.
f(4+h) -f(4) = ((4+h) / ((4+h)^2 +1))) - 4 / (4^2 + 1)

Could someone help me expand this and give me some derivation tips?

P.S Thx to your help I got a B+ on my last math thest which I've never had before! Thanks ALOT!
D.S

Here I will go thorough the definition of the derivative for the first one and see if it helps...ok we have $\displaystyle f(x)=3x^4-350x+129$....and we have $\displaystyle f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$...so we have $\displaystyle f'(x)=\lim_{h \to 0}\frac{3(x+h)^4-350(x+h)+129)-(3x^4-350x+129)}{h}$...expanding all but the first term we get $\displaystyle f'(x)=\lim_{h \to 0}\frac{3(x+h^4)-350x-350h+129-3x^4+350x-129}{h}$...simplifying we get $\displaystyle f'(x)=\lim_{h \to 0}\frac{3(x+h)^4-350h+3x^4}{h}$./...now expanding we get $\displaystyle f'(x)=\lim_{h \to 0}\frac{3x^4+12x^3h+18x^2h^2+12xh^3+3h^4-350h-3x^4}{h}$..combining like terms we get $\displaystyle f'(x)=\lim_{h \to 0}\frac{12x^3h+18x^2h^2+12xh^3+3h^4-350h}{h}$....now by splitting the top apart we get $\displaystyle f'(x)=\lim_{h \to 0}\bigg[12x^3+18x^2h+12xh+3h^3-350\bigg]$...now since all the terms with h go to zero just the terms with just x are left...therfore $\displaystyle f'(x)=12x^2-350$...and then $\displaystyle f'(4)=12\cdot{4^3}-350=418$
• Apr 12th 2008, 06:13 AM
Hanga
Yeah I can see that there is no shortcut in this :P
Could you guys help me with b) please? I'd really like to see how you solve that one.
• Apr 12th 2008, 06:18 AM
Moo
Ok :P

$\displaystyle f(x) = \frac{x}{x^2 + 1}$

$\displaystyle \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$ ?

First of all, do you know the other formula :

$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ ?
• Apr 12th 2008, 06:20 AM
Hanga
Quote:

Originally Posted by Moo
Ok :P

$\displaystyle f(x) = \frac{x}{x^2 + 1}$

$\displaystyle \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$ ?

First of all, do you know the other formula :

$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ ?

Yeah I know the
$\displaystyle \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$

I do not know this one
$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ ?[/quote]
Simply because there is no example on how to execute this in my lousy **** book...

enlighten me S.V.P :)

I tired using this;

(f(4+h) - f(4) / h)
• Apr 12th 2008, 06:25 AM
Mathstud28
I'll let moo take the work
Quote:

Originally Posted by Hanga
Yeah I know the
$\displaystyle \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$

I do not know this one
$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ ?

Simply because there is no example on how to execute this in my lousy **** book...

enlighten me S.V.P :)

I tired using this;

(f(4+h) - f(4) / h)[/quote]
but the second one is the alternate form of the deriavative that says f'(x) evaluated at c is $\displaystyle f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}$
• Apr 12th 2008, 06:33 AM
Hanga
Quote:

Originally Posted by Mathstud28
Simply because there is no example on how to execute this in my lousy **** book...

enlighten me S.V.P :)

I tired using this;

(f(4+h) - f(4) / h)

but the second one is the alternate form of the deriavative that says f'(x) evaluated at c is $\displaystyle f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}$[/quote]

Here is what I don't undestand. How do I use the $\displaystyle f'(c)=\lim_{x \to c}\frac{f(x)-f(c)}{x-c}$ Can you please show me an example?

For me this looks like it's always going to be 1....
• Apr 12th 2008, 06:35 AM
Moo
Quote:

Originally Posted by Hanga
Yeah I know the
$\displaystyle \text{What \ is \ } \lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$

I do not know this one
$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ ?
Simply because there is no example on how to execute this in my lousy **** book...

enlighten me S.V.P :)

I tired using this;

(f(4+h) - f(4) / h)

What a pity >_<

Then you should know the one i gave you, it'll be easier to calculate i think.
You can get this formula by changing variables :
h -> x-a

That is to say h+a -> x

Forget it , i'm not sure it's more simple
• Apr 12th 2008, 06:44 AM
Hanga
Quote:

Originally Posted by Moo
What a pity >_<

Then you should know the one i gave you, it'll be easier to calculate i think.
You can get this formula by changing variables :
h -> x-a

That is to say h+a -> x

Forget it , i'm not sure it's more simple

Please give me an example using
$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$

what is the difference between f(a) and a? What is the diffirence between f(x) and x?

If I take f(x) - f(a) / x - a

wont the sum alwas become 1?

f(x) - f(2) / x - 2 = 1!?

please just give me an example so I can see how you use this formula
$\displaystyle f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$

• Apr 12th 2008, 06:45 AM
Moo
Ok, i went through it.

I'm sorry, but it's far less complicated to write it than type it ;)

If you have questions about it, just tell me which line !
• Apr 12th 2008, 06:51 AM
Hanga
thanks mate, thats awsome!

I understand how to do something like that now :)

Viva le france :)
• Apr 12th 2008, 06:58 AM
Moo
You're welcome :p

And f(x) is completely different from x !
f(x) is the image of x by the function f. It will not necessarily return the same value :o)