• Apr 22nd 2010, 04:06 AM
j55
hi, ive just been reading through this website on differentiation, anyway ive just come across subtraction and addition of functions, KryssTal : Introduction to Calculus theres the link, if you scroll down about 1/4 of the way down you will see the section i am talking about, can anyone explain this a bit easier i cant seem to grasp the concept, especially when i was following an example; (i) For the function y = 4x2 + 2x + 3, the derivative is dy/dx = 8x + 2
now i cant see where the 2 has come from, i get the 8x part but not the second bit.
• Apr 22nd 2010, 04:32 AM
Dint
Hi,

if you know that for
$y(x) = ax^n$

$dy/dx = a n x^{n-1}$

think about $y(x) = 5x$

here a = 5 and n = 1

now do the substitution with the formula I gave you...
• Apr 22nd 2010, 04:34 AM
craig
Is this you first your first experience with Calculus?

To differentiate basic functions such as the one above, you multiply the coefficient by the power, and then reduce the power by one.

For example:

The derivative of $4x^2$ is $(4 \times 2)x^{2-1} = 8x^1 = 8x$

As for the second bit:

$2x = 2x^1$, so differentiating this gives you $(2 \times 1)x^0 = 2(1) = 2$.

Hope this clear is up a bit. A great place to check out if you want to learn the basics of calculus is Khan Academy on Youtube.

• Apr 23rd 2010, 11:41 AM
j55
re
yeah it does help, thanks. just a couple of queeries though..

so if in such a problem if theres an x with no power then the outcome is going to be just a number without the x ? also you ended up with 2(1) where did you get the 1 from?

and yeah it is my first experience hence the uncertanties.

also in the origional post, y = 4x2 + 2x + 3 now where does the 3 go?
• Apr 24th 2010, 12:41 PM
craig
Quote:

Originally Posted by j55
so if in such a problem if theres an x with no power then the outcome is going to be just a number without the x ? also you ended up with 2(1) where did you get the 1 from?

Yes it would just be the constant without the x. I got the 1 because $x^0 = 1$

Quote:

Originally Posted by j55
also in the origional post, y = 4x2 + 2x + 3 now where does the 3 go?

If you differentiate a constant it goes to zero, you could think of it like this:

$3 = x^0$, so differentiating it would mean that you multiply by zero.
• Apr 25th 2010, 02:35 AM
j55
re
thats great thanks for your help craig much appreciated