i get it and i don't get it...

Quote:

Originally Posted byxpi0t0s> Now, when you differentiate f(x) : [[ D(h(x)) = Derivative of h(x) w.r.t. x ]]

> you do it like f'(x) = D(x) + D(x) + ..(x times).. + D(x) which is invalid.

Why is it invalid? According to http://mathworld.wolfram.com/Derivative.html "Derivatives of sums are equal to the sum of derivatives" (exact quote), see equation 36: [f(x)+...+h(x)]' = f'(x)+...+h'(x).

Did you go through the next line : D(nx) = nD(x)

**ONLY**for constant value of n.

You can say D(x + x + x) = 3.

But, D(x + x + ..(y times).. + x) != y when y depends on x.

When y depends on x, D(x + x + ..(y times).. + x) = yD(x) + xD(y).

You

**have to**apply product formula 'coz you are expressing one function (in this case, x) as the sum with itself, another function (in this case, y) times.

In your question, y depends on x as y=x.

So, f'(x) = xD(x) + xD(x) = 2x != x.

And,

**LOL**@ what you mentioned as the correct answer.

You are trying to show that 2x != x by substituting x = 4 !!

The first principle and blah.. blah.. blah.. was simply

**unnecessary**. Yeah it was unnecessary. You get the same results by differentiation formula too and you can still show x != 2x by showing 4 != 8.

**LOL**

I wonder the basis on which my answer was rejected. The correct answer (by the OP) sounds too silly.

And, one more thing; the following (in bold) is the silliest thing I have ever seen in my life :

Quote:

Originally Posted byxpi0t0sIf f(x)=f1(x)+f2(x)+f3(x)+...+fN(x) then by equation 36:

f'(x)=f1'(x)+f2'(x)+f3'(x)+...+fN'(x)

So if fp(x)=x (where p=1..N) and fp'(x)=1 then

f'(x)=1+1+1+...1 (N times)

so f'(x)=N.

If N=x then f'(x)=x.

**very day**.

(By construction this must be a positive integer, of course)

As I gave two examples where it does hold I think that's fair enough.

Let me repeat those two examples just in case it's not clear:

Let's say x=3. f(x)=x+x+x; f'(x)=1+1+1=3 so f(x)=x.

x=7: f(x)=x+x+x+x+x+x+x; f'(x)=1+1+1+1+1+1+1=7 so f(x)=x.

So let's see a counter-example rather than just a declaration that I'm silly.

Let's say x=3. f(x)=x+x+x; f'(x)=1+1+1=3 so f'(x)=x.

x=7: f(x)=x+x+x+x+x+x+x; f'(x)=1+1+1+1+1+1+1=7 so f'(x)=x.

this is gonna be controversial....

Things have been sorted out now ....

Let's get back to the topic then.

Just go to some maths prof, and if he agrees that f'(x) = x is true, I will truly give up maths and programming. Just announce it here that the prof agrees that f'(x) = x.

And, what counter-example are you talking abt ?

Differentiation is not a game or something -- calculus is a vast branch of Maths and what you are trying to show will prove Newton and Leibniz's work false !!

Quote:

Originally Posted byxpi0t0sLet's say x=3. f(x)=x+x+x; f'(x)=1+1+1=3 so f(x)=x.

x=7: f(x)=x+x+x+x+x+x+x; f'(x)=1+1+1+1+1+1+1=7 so f(x)=x.

**big**misconception of yours.

Let me talk your way :

Let's have x = 3.

So, f(x) = x + x + x = 3 + 3 + 3 = 9

=> f'(x) = 0.

Agree ??

Same for x= 7 or anything. Now, do you agree to this ???

Please do seriously revise differentiation, b4 next post. And, I am serious abt this thing, not 'coz it's some contest question or anything; but 'coz I

**love**maths.

And, I do stick to facts ... but not false ones.

And I can very boldly and (may be rudely) oppose those who support

**any**false facts.

Let's set N=5 and so f(x)=x+x+x+x+x - that's 5 X's, because N=5.

Let's also completely ignore the value of x for now.

Now let's differentiate f(x); this gives 1+1+1+1+1, which is 5.

We can further simplify f(x)=5x and this differentiates directly to 5, because d/dx (ax^b) = abx^(b-1).

Do you at least agree with that?

'coz N is independent of x and hence can be treated as a constant.

Do, D(Nx) = ND(x).

But you can't do so when N depends on x.

Like, you can't take N = x.

Like, you can't take N = x.