Why does the derivate of (x+1)^2 not equal 2?

Question

$$f(x) = (x+1)^2$$

$$f'(x) = 2(x+1)$$

Shouldnt it equal 2 because, the rules of derivation says that

$$(x+1)^2 =x^2+1$$

$$1=0$$

$$x^2=2x=2$$

$$2+0=2$$

looks like this site doesn't have empathy with newbies - just made a proof of that :) — user3025060, Dec 20 '13 at 18:09
From what I know the chain rule tells me that $f'(x)=2(x+1)'(x+1)^{2-1}$ — Module, Dec 20 '13 at 18:52
@Module I used another method to calculate the derivative, your rule can be used. what I did was removing exponet and then calculating the derivation using the method above. — user3025060, Dec 20 '13 at 18:57
You had an error when foiling it. Always remember what you've learned in algebra: $(a+b)^2=a^2+2ab+b^2$ — Module, Dec 20 '13 at 19:00
BTW,. I study Maths in another language, so if something was translated wrong, here's what I mean: 1)Exponet: what appers above the function (e.g $$x^2$$) 2)calculating the derivation using the method above: I mean after getting rid of the exponet, I used the normal derivatives rules (e.g $$A$$ = 0, $$Ax = A$$ ..etc) just to not be understood incorrectly :) — user3025060, Dec 20 '13 at 19:04
@Module I always forget the basics, this is always causing me troubles :-( — user3025060, Dec 20 '13 at 19:06
I understood exactly what you said, but if you want to further develop you need to get the basics down. — Module, Dec 20 '13 at 19:11

score 5 · Accepted Answer · answered Dec 20 '13 at 17:57

5

Your mistake is not calculus related: $$(x+1)^2 = x^2 + 2x + 1$$

You have (incorrectly) written that: $$(x+1)^2 = x^2 + 1^2$$

Thus, you can easily see the derivative should be $2x+2$.

answered Dec 20 '13 at 17:57

apnorton

stupid me! I forgot something like that - thanks – user3025060 Dec 20 '13 at 17:58

1 Answers1