anti-derivative of f'(x) = 0

Question

anonymous · Answer

@beccaboo333

beccaboo333 · Answer

I have no idea how to do this kind of math.. I'm sorry :c

anonymous · Answer

it's okay :)

amistre64 · Answer

what derivaitve rule gets us a derivative of 0?

anonymous · Answer

it's chain right?

amistre64 · Answer

in general ... no

anonymous · Answer

Could you hint me? i'm a little bad with calc.

amistre64 · Answer

the hint will be:

tell me the derivative rules that you rememebr

anonymous · Answer

chain rule
quotient rule
product rule

amistre64 · Answer

exponent rule, log rule ... none of those say f'(x)=0

amistre64 · Answer

the most basic rule is for something of the form:  f(x) = x^n

anonymous · Answer

the part that confuses me is "f'(x)" = 0 because the antiderivative of 0 is 0 + c.

amistre64 · Answer

they split this rule into two forms simply becuase they dont like x^0 for x=0, otherwise its the same

amistre64 · Answer

and 0+c is just some constant

anonymous · Answer

oh I was off with that. :/

amistre64 · Answer

$$y=x^0$$$$y'=0x^{-1}=0$$
but if x=0 then we get a divide by 0 that everyone frowns upon so they simply make the rule that x^0 is some constant, or just 1  such that:

$$y=kx^0$$
$$y=k*1$$
$$\frac{d}{dx}(y)=\frac{d}{dx}k*1$$
$$\frac{d}{dx}(y)=k*\frac{d}{dx}1$$
$$y'=k*0$$

anonymous · Answer

so the final turns out to be k * 0, not 0 + c right?

amistre64 · Answer

0+c is fine,  or k+c, or just C

anonymous · Answer

f'(x)= k * 0

anonymous · Answer

tks

amistre64 · Answer

yep