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 ?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
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
@plasmataco this is calculus ii material
you can see taking a derivative is a linear operator if you write out the first principles
A moderator is present in this conversation. Do not go off topic and break the rules
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..
I gave mine to dan, so... yeah, give to @hartnn
mine is also given to dan
Its been years i didn't care about medals and stuff, so feel free to give it to anyone you like :)