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I'm trying to teach myself algebra and derivatives. I learned the derivative for $f(x) = x^2$ from a lesson, and now I thought I would see if I could figure out the derivative of $f(x) = x+x$ on my own.

I know the formula for derivatives is: $$\lim_{\delta\rightarrow 0}\frac{f(x+\delta) - f(x)}{\delta}$$

So my attempt at algebra amounted to this: $$\frac{((x+x) + (\delta + \delta)) - (x+x)}{\delta}$$

$$=\frac{\delta + \delta}{\delta}$$

Which doesn't seem right. (Isn't the derivative supposed to NOT contain the delta term?)

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That's right, after you take the limit, you should have no more $\delta$'s in the expression. However, you have not yet taken a limit in the work you show above. –  wckronholm May 28 '11 at 16:16
    
@wckronholm: I thought derivatives allow you to calculate slope without taking a limit? –  Matt Munson May 28 '11 at 16:23
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NO! The derivative IS a limit. There are some short-cuts for computing derivatives which you'll learn at some point which will suppress the limit computations, but this doesn't mean that the derivative is no longer a limit. –  wckronholm May 28 '11 at 16:26
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Also, this particular computation might go smoother if you simplify $x+x=2x$ before you start computing the derivative. –  wckronholm May 28 '11 at 16:30
    
Also, note that $f(x+\delta)$ where $f(x)=x+x$ should be $x+\delta + x+\delta$. It is equivalent to what you wrote, but maybe you just got lucky? –  Vhailor May 28 '11 at 23:55
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2 Answers

up vote 10 down vote accepted

Add the $\delta$s and simplify, you get $\displaystyle \frac{2\delta}{\delta} = 2$. The derivative is then $2$.

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I thought about that, but if delta is 1, 1*1/1 = 1. Also, can the derivate really be 2? doesn't the slope of this function increase with x? And I know the derivative can't be 2 for x=0. –  Matt Munson May 28 '11 at 16:21
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Part of the point of these exercises is to realize that $\frac{2\delta}{\delta}=2$ only if $\delta\neq 0$. This is where the limits come in. –  wckronholm May 28 '11 at 16:23
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Err... Isn't it a $+$ between the two $x$s? Then you get $x+x=2x$ which is a straight line, its derivative is constant. –  Najib Idrissi May 28 '11 at 16:23
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yeah, its a +, but I though 2x is the same as saying x*x... Wait...DOH! haha yeah I see what you mean. And I also now see that it does graph a straight line. Thanks –  Matt Munson May 28 '11 at 16:30
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Here is another great way to calculate the derivative of $f(x)=(x+x)$, instead of using the Limit Definition, which is the method you are using above.

The other way to do this is to use a combination of methods called the Power Rule, and Sum Rule.

Here is a mathematical representation of the Sum Rule:

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

This says $(f + g)$ prime, where prime means "the derivative of." So this rule literally states: the derivative of $(f + g)(x)$ is equal to the derivative of $f$, plus the derivative of $g$.

In your example this would break down to:

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

Now we will use the Power Rule to calculate the derivative of each individual piece (the $x's$ in this case).

The Power Rule states:

$$\frac{d}{dx} x^n = nx^{n-1} $$ Note: $\displaystyle \frac{d}{dx}$ means to take the derivative with respect to $x$. So this rule states that the derivative of $x^n$ is equal to multiplying $x$ by $n$,and then subtracting $1$ from the exponent.

Here is the Power rule applied to $x' + x'$:

Remember that $x$ is really $x^1$, so

$$\frac d{dx}(x^1) = 1\cdot x^{1-1} = x^0 = 1$$

Applying this rule to both $x's$ we end up with

$$1 + 1 = 2$$

Which is your same result. Hopefully this made sense you. If you get the hang of these rules they make calculating your derivatives super quick, and also provide you with a way to check your Limit Definition calculation. Good luck!!

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The power rule comes from two things (well, more if you are taking the power not to be a positive integer): the product rule, and the calculation of $D_x(x)=1$. When you are just starting calculus, using the power rule for this particular derivative is a bad thing. You are replacing "this is why it is true" with "here is a formula that gives the right answer, and which, in lieu of the calculation we have replaced, cannot be verified." Yes, it is important to have formulas, but it is more important that everything is well grounded. –  Aaron May 28 '11 at 23:40
    
@Aaron: I agree. I edited the answer directly above to be more explicit about when Chris was taking a derivative, so as not to confuse the person posing the question. In particular: prior to editing, the answer above asserted: $x^1 = 1\cdot x^{1-1} = x^0 = 1$; essentially asserting that "$x = \frac d{dx}(x)$". –  amWhy May 28 '11 at 23:53
    
Thanks for the feedback. I always learned, solve first, prove later. It made it more conformable when learning the formal "reasons why," after having already applied the formulas in practice. @amWhy: Thanks for the clarification –  Chris Camargo May 29 '11 at 0:08
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@Chris Camargo Well, I don't in general have a problem with learning how to calculate more complicated things before (or sometimes instead of) knowing exactly why things work. To a certain extent I wouldn't mind if my students didn't calculate the derivative of $x^2$ from the definitions. I just think that this one derivative (and the derivative of a constant) need to be done and understood before you move on to calculational tools. Take everything else in calculus on faith, but leave me those two calculations. –  Aaron May 29 '11 at 0:59
    
@Chris: no problem. I knew you knew what you were doing, and what you meant by what you wrote, but I didn't want the person posting the question to get confused. –  amWhy May 29 '11 at 1:25
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