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

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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**.

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(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.

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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.

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this is gonna be controversial....

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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.

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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?

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

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

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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.