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# Derivative Properties

The definition of the derivative does what we need it to do but it isn't the easiest tool to use. This often happens in math when we're trying to extend an existing idea (the sloep of a line, in this case) rather than just coming up with something new. In this situation, one of the first things we do is look for simplified methods and formulas that, while they might not apply in every case, can help with ones that are common. Two of the formulas that we're going to see in the next lecture are:

$$\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))$$

and

$$\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$$

The first one says that if we have a function that's got a constant multiple in front, e.g. $f(x) = 3x^2$, then we can just find the derivative of the $x^2$ part and multiply the result by the 3. The second rule says that if we want to find the derivative of the sum of two functions, we can just find the derivatives of the individual parts and add those together.

We're going to see the proofs of these properties in the lecture but they're something you should try to work out on your own first. You can walk through the derivation of the first one by cicking on the boxes covering the individual steps below.

Show that $\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x)$.

 Start with the definition of the derivative as a function $$f'(x) = \lim_\limits{h\to 0}\frac{f(x + h)-f(x)}{h}$$ Next replace $f(x)$ with $cf(x)$. $$\frac{d}{dx}(cf(x)) = \lim_\limits{h\to 0}\frac{cf(x + h)-cf(x)}{h}$$.\ Now factor out the c $$\frac{d}{dx}(cf(x)) = \lim_\limits{h\to 0}\frac{c(f(x + h)-f(x))}{h}$$ Now bring the c outside of the limit. $$\frac{d}{dx}(cf(x)) = c\lim_\limits{h\to 0}\frac{f(x + h)-f(x)}{h}$$ Apply the definition of the derivative to what's left in the limit. $$\frac{d}{dx}(cf(x)) = c \frac{d}{dx}f(x)$$

The proof of the second statement is very similar, you just have to rearrange the terms in the numerator so the f's and g's are together then split the limit in two.