Math Is Fun Forum

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivatives

You are still using the old formula, you don't need that anymore.  For instance if you wanted to differentiate x^6 you could apply the power rule and get 6x^5. If you wanted to diff, 32 x by the  rule  D ( c * x ) = c you would get 32. Here is  rule for differentiating just a constant:

So D (  12 ) = 0. With these 3 rules you can get the derivative for the ball problem easily, try it.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Derivatives

Ah, I think I'm getting it now.

So taking the following: f(t)= -16t^2+100t
f'(t) = 2(-16)(t)^1 + 1(100)(t)^0
f'(t) = -32t + 100

So the exponent is only applied to the variable (in our case 't') and not the constant, right?

Last edited by Anakin (2009-12-12 19:07:57)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivatives

Yes, only to the variable. Memorize those three rules and here are more:

You are intersted in differentiation of simple functions, about 1/2 down the page.

http://en.wikipedia.org/wiki/Table_of_derivatives

You will eventually memorize most of those, then differentiating functions will be much easier.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Derivatives

Wow, that is REALLY cool. Quite simplified (at least for these basic functions).

Just to reconfirm to make sure I'm taking the right message home: For the derivative, the co-efficient is the power of the term multiplied by the original co-efficient, then all that is multiplied by the original variable to the power of n-1 (where n was the original power) to get the derivative.
Ex.
f(x)=4x^3-2x^2
f'(x) = 3 (4)(x)^2 - 2(2)(x)^1
f'(x) = 12x^2 - 4x
That is the correct concept behind what we discussed today, as to my understanding now. 

And, I looked at that link, and tried all the functions without looking at the answer, and got them all!

Last edited by Anakin (2009-12-12 19:34:24)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivatives

Yes, that is correct! Hold on to that page for reference. One last simple problem :

Differentiate sqrt(x)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Derivatives

f(x)=sqrt x
f(x) = x^0.5
f'(x) = 0.5(x)^-0.5
f'(x) = 0.5(1/x^0.5)
f'(x) = 0.5/sqrt(x)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivatives

Morning;

Thats correct. When possible try to work in fractions rather than in decimals. Otherwise you will have problems with x^(1 / 3).

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Anakin

Member

Registered: 2009-10-04

Posts: 145

Re: Derivatives

Alright, will do!