# Why is the derivative of $x^2$ not $2x+1$?

If the derivative is the change of the function at each step, it could be expressed as:

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

Therefore if $f(x)=c$

$$c+f'(x)=c \implies f'(x)=0$$

This is also correct for $f(x)=cx$

$$cx+f'(x)=c(x+1) \implies f'(x)=c$$

However, it doesn't work for $f(x)=x^2$

$$x^2+f'(x)=(x+1)^2=x^2+2x+1 \implies f'(x)=2x+1$$

Where am I wrong in my reasoning?

• derivative of a function is not change in the value of the function. it is a rate(ratio/quotient/fraction) of change of a function over smaller and smaller change in $x$ at a particular point. – abel Dec 10 '14 at 11:10
• I disagree with the downvote. Of course there is an error in the given reasoning, but the question is precisely to spot this error. Moreover, the question "shows research and effort", surely is "clear" and probably it might be also "useful" for someone studying the basics of differential calculus. IMHO, it deserves to be upvoted. – Giuseppe Negro Dec 10 '14 at 11:15
• @GiuseppeNegro The question didn't have LaTeX. It was very hard to read before I edited it. But I am not the downvoter. – Mark Fantini Dec 10 '14 at 11:17
• @MarkFantini I don't know when the failure of a new user to know how to use, and use, mathjax was grounds for a downvote. It is reasonable to expect someone whose asked a handful of questions to learn and use it, and ditto for answering. But be reasonable and forgiving and try to be more welcoming to newcomers! – Namaste Dec 10 '14 at 12:11
• Thanks, I will make sure to format my question correctly next time. As you might have guessed I'm not a mathematician. I am in fact a programmer. There are always exceptions, but in general this site seems more welcoming than stackoverflow. – user2944397 Dec 10 '14 at 14:57

The thing you take for a derivative is $$\Delta_1f(x)=f(x+1)-f(x).$$ Similarly, you can define an analogous thing for different step sizes $h\neq0$: $$\Delta_hf(x)=\frac{f(x+h)-f(x)}{h}.$$ The derivative is defined as the limit of these difference quotients as $h\to0$: $$f'(x)=\lim_{h\to0}\Delta_hf(x).$$ Let's consider your concrete example, $f(x)=x^2$. For it we can calculate that $\Delta_hf(x)=2x+h$, so that $\Delta_hf(x)\to2x$ as $h\to0$. In short, the main point of this answer is that typically $\Delta_1f(x)\neq f'(x)$.

• Very clear explanation. Thanks. – user2944397 Dec 10 '14 at 15:40

You are confusing discrete functions, that is, functions which are only defined for integers or natural numbers, with continuous functions on the real numbers.

The derivative of a function like $f(x)=x$ on all the real numbers is NOT defined as the change of the function at each step because there are no such things as steps.

If, however, you only consider discrete functions and actually call the increment of the function at each step its derivative, you are right when you say that the derivative of $f(n)=n^2$ is $2n+1$.

Where are you wrong?

$x^2 + f′(x)=(x+1)^2$
The rate of change is itself changing. In order to get from $f(a)=a^2$ to $f(a+1)=a^2+(2a+1)$, earlier (closer to $x=a$) the $x^2$ function is changing more slowly than when you get closer to $x=a+1$. Overall the average rate of change is $2a+1$, but instantaneously, at $a$ the rate of change is $2a$, while at $a+1$ the rate of change is $2a+2$.