proof for calculus?

Question

how do you get the proof that f(X) = 3 and its derivative is 0, zero. I mean its obvious that its a constant and its derivative is always 0, but how do you prove it?

hint: by using (f(x + dx) - f(x) )/ dx as lim dx goes to 0

Answer 1

f(x) = 3
f(x + h) = 3 where h = dx

f `(x) = [ f(x + h) - f(x) ] / h---------h ≠ 0
f `(x) = [ f(x + h) - f(x) ] / h
f `(x) = [ 3 - 3 ] / h
f `(x) = 0

Answer 2

It is much easier to use the definition directly, but usually when in a curriculum the power rule is proven first by this method and then it is used to prove derivative of a constant.

lim(h->0) [f(x+h)-f(x)]/h given that f(x) = ax^b
lim(h->0) [a(x+h)^b - ax^b]/h
Concentrating on the a(x+h)^b and expanding it, we notice that when using the binomial expansion thereom (google it), the first term will be ax^b and the last will be ah^b. The first term always adds with the -f(x) afterwards, and we are left with a numerator where every term has an h in it. We factor out an h and cancel top h and bottom h. That leaves us with abx^(b-1) + (terms with h) that when fill in h with zero we get just that term. That is (roughly) the proof for the power rule. Written out ~ f(x) = ax^b f'(x) = abx^(b-1)

So for any constant function
f(x) = c = cx^0 f'(x) = 0* x^(-1) = 0

Answer 3

Ok. Left F(x)=c. Then, the process you said.
d/dx[c]=f'(x).
=lim as h-0 (f(x+h) - f(h))/(h)
=lim as h->0 (c-c)/(h)
The C's cancel out, making 0 a numerator.
=Lim as h->0 = 0.
So, when a constant is there, the lim as it approaches 0 is 0.

Answer 4

f(x+dx) = 3
f(x) = 3
f(x+dx) - f(x) = 0
No matter what the denominator does as dx --> 0, the numerator is 0; hence, f'(x) = 0

Answer 5

f(x) = 3

multiply RHS with x/x

f(x) = 3x/x

using product rule

f'(x) = [x*3 - 3x*1]/x^2 = (3x - 3x)/x^2 = 0