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because derative is a linear operator
constants can be factored out of the derivative
wow this guy must be rich to do a QH question
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
no just desperate, lol .. I've given up caring about money
so I just didnt simplify it enough?
no its not like that its just that the derivative is a linear operator
you can see taking a derivative is a linear operator if you write out the first principles
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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 .
/2 is division by 2. multiplication or division of a constant to the function, multiplies or divides its derivative by the same constant.. \(\Large if ~ ~f'(x) = g(x), \\ then , ~~ [a \times f(x)]' =a \times g(x) \)
ah ok.. that's the piece I was missing hartnn
even though its a /2 .. it's a constant multiple.
ah ok.. that makes things click for me.
sorry dan.. that was a colossal effort, much appreciated.. you lost me a bit there though :)
thats all there is to it, i was showing you why, you can factor the constant out
that's why he's the honorary professor of Mathematics XD
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..
mine is also given to dan
thanks hugh! Its been years i didn't care about medals and stuff, so feel free to give it to anyone you like :)