If the derivative of
f[x_] = (x^2 - 2 Log[x])
f'[x_] = (2x - 2/x)
Then why is the derivative of
f[x_] = (x^2 - 2 Log[x])/2
f'[x_] = (2x - 2/x)/2
What is the principle that allows the /2 which appears to be some kind of constant to remain in the derivative ?

- anonymous

- schrodinger

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- dan815

because derative is a linear operator

- hartnn

constants can be factored out of the derivative

- UsukiDoll

wow this guy must be rich to do a QH question

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## More answers

- dan815

f[x_] = 2*(x^2 - 2 Log[x])
heres another example this is same as
f[x_] = (x^2 - 2 Log[x]) + (x^2 - 2 Log[x])
there are 2 of them, and if u remember we can differentiate each term separately

- anonymous

no just desperate, lol .. I've given up caring about money

- anonymous

so I just didnt simplify it enough?

- dan815

no its not like that its just that the derivative is a linear operator

- UsukiDoll

@plasmataco this is calculus ii material

- dan815

you can see taking a derivative is a linear operator if you write out the first principles

- UsukiDoll

A moderator is present in this conversation. Do not go off topic and break the rules

- dan815

|dw:1434803107603:dw|

- UsukiDoll

maybe we can try to take the derivatives together.. one is straight forward
first it's just derivative and then second is a simple product rule .

- dan815

|dw:1434803242178:dw|

- hartnn

/2 is division by 2.
multiplication or division of a constant to the function,
multiplies or divides its derivative by the same constant..
\(\Large if ~ ~f'(x) = g(x), \\ then , ~~ [a \times f(x)]' =a \times g(x) \)

- anonymous

ah ok.. that's the piece I was missing hartnn

- anonymous

even though its a /2 .. it's a constant multiple.

- hartnn

exactly!

- anonymous

ah ok.. that makes things click for me.

- anonymous

sorry dan.. that was a colossal effort, much appreciated.. you lost me a bit there though :)

- dan815

thats all there is to it, i was showing you why, you can factor the constant out

- UsukiDoll

that's why he's the honorary professor of Mathematics XD

- dan815

lol :>

- anonymous

I got to give this one to hartnn though.. that was the piece I was missing.. I knew the /2 had to carry into the derivative, but I could not work out for the life of me, what rule was making that happen.. once I realized it was just a 0.5 f[x] constant multiple, then it clicked..

- Plasmataco

I gave mine to dan, so... yeah, give to @hartnn

- UsukiDoll

mine is also given to dan

- hartnn

thanks hugh!
Its been years i didn't care about medals and stuff, so feel free to give it to anyone you like :)

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