Derivative of $f(x) = (x+x)$

Solution 1:

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

Solution 2:

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!!