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 ?

Question

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  ?

dan815 · Answer

because derative is a linear operator

hartnn · Answer

constants can be factored out of the derivative

usukidoll · Answer

wow this guy must be rich to do a QH question

dan815 · Answer

​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 · Answer

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

anonymous · Answer

so I just didnt simplify it enough?

dan815 · Answer

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

usukidoll · Answer

@plasmataco this is calculus ii material

dan815 · Answer

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

usukidoll · Answer

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dan815 · Answer

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