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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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    @wckronholm: Aren't we supposed to let $\delta$ be more closely to zero. So, by this assuption, he couldn't take "1" for it as he did in his comment below? However this question was answered so completely.2012-01-17

2 Answers 2

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

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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. Thanks2011-05-28
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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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    @Patrick Da Silva I agree with you. I just wanted to give Matt some insight to other possible ways of solving this problem.2011-05-29