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+.EQ
+delim $$
+.EN
+.TL
+Derivative Identities
+.AU
+Preston Pan
+.AI
+Pacific School of Innovation and Inquiry
+
+.NH 1
+Introduction
+.PP
+Now that you know how to calculate the derivative of a specific very simple function,
+you might now ask how you might take a derivative of a more complicated function, perhaps
+involving functions that are the sum or product of two other simpler functions, or the
+composition of two functions.
+
+.PP
+In order to get to a general case, we can look at a specific case. Say, for example,
+we want to take the derivative of $f(x) = x sup 2 + x$ (we are using h for delta x here because
+that is the actual convention):
+
+.EQ
+f'(x) = {{(x + h)} sup {2} + x + h - {x} sup {2} - x} over h
+.EN
+.EQ
+f'(x) = {{x} sup {2} + 2xh + {h} sup {2} + x + h - {x} sup 2 - {x}} over h
+.EN
+.EQ
+f'(x) = {2xh + {h} sup {2} + h} over h
+.EN
+.EQ
+f'(x) = 2x + 1 + h
+.EN
+
+.PP
+and as h becomes infinitely small, the resulting derivative is 2x + 1.
+
+.PP
+But we know already that the slope of x was equal to one. You learn that in 9th grade.
+And we know that 2x is the derivative of $x^2$. So it seems like this should be true:
+
+.EQ
+(f + g)' = f' + g'
+.EN
+
+.PP
+or in words: adding the functions and then taking the derivative is exactly the same as taking
+the derivative of both the functions first then adding them. In other words: the order of adding
+and taking the derivative doesn't matter. But is this really true?
+
+.PP
+In fact, it is! If we use two general functions f and g, we can see that this is true for any
+two functions that you pick:
+
+.EQ
+(f + g)' = {f(x + h) + g(x + h) - f(x) - g(x)} over h
+.EN
+
+and if we just rearrange and separate the g and f terms:
+.EQ
+(f + g)' = {f(x + h) - f(x)} over {h} + {g(x + h) - g(x)} over h = f' + g'
+.EN
+
+If that is not a clear illustration, this is extremely easy to figure out on your own given
+the general method.
+
+It should be clear that multiplication can be done in the same general process. However,
+it is a little bit more complicated. I suggest trying to figure it out on your own before
+you read the solution below:
+
+.EQ
+(f * g)' = {f(x + h)g(x + h) - f(x)g(x)} over h
+.EN
+in order for this solution to work, we must subtract and add a term $f(x + h)g(x):$
+.EQ
+(f * g)' = {f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x)} over h
+.EN
+and we factor some terms out:
+.EQ
+(f * g)' = {f(x + h)(g(x + h) - g(x)) + g(x)(f(x + h) - f(x))} over h
+.EN
+and we can clearly see that:
+.EQ
+(f * g)' = f(x + h){g(x + h) - g(x)} over {h} + g(x){f(x + h) - f(x)} over {h}
+.EN
+as $h$ approaches zero, $f(x + h)$ approaches $f(x)$. Also, we can see that some of
+these terms look like derivatives, so:
+.EQ
+(f * g)' = f(x)g'(x) + g(x)f'(x).
+.EN
+
+.PP
+And this will work with any two functions where you know their derivatives. Isn't that cool?
+
+.NH 1
+The power rule
+.PP
+Up until now, we assumed that you could take the derivative of an arbitrary function $f$ and
+$g$ and gave rules for computing the derivatives of their products and sums based on that
+assumption. However, it's not clear how you are supposed to just
+.I "know"
+the derivatives of many functions, including sine and cosine, as well as $x sup n$. Of course,
+we figured it out for $x sup 2$, but there are many functions that we have not explained the
+derivative of. How do we find these derivatives?
+
+.PP
+Of course, like with all derivatives of functions, you can calculate them with the general
+derivative definition. Here, we will discuss the power rule, or $x sup n$ for any positive
+integer $n$.
+
+.PP
+If we just plug it into the general form form directly:
+.EQ
+f'(x) = {{(x + h)} sup {n} - {x} sup {n}} over h
+.EN
+You might observe that we need to somehow expand the binomial ${(a + b)} sup n$ for arbitrary n.
+You might try doing this by expanding for $n = 1$, $n = 2$, etc... and finding a pattern:
+.EQ
+{(a + b)} sup 0 = 1
+.EN
+.EQ
+{(a + b)} sup 1 = a + b
+.EN
+.EQ
+{(a + b)} sup 2 = a sup 2 + 2ab + {b} sup {2}
+.EN
+.EQ
+{(a + b)} sup 3 = a sup 3 + 3{a} sup {2}b + 3{b} sup {2}a + {b} sup {3}
+.EN
+and if you keep on doing this for higher $n$, you will see that:
+.EQ
+{(a + b)} sup n = a sup n + n{a} sup {n - 1} {b} sup {1} + ... b sup n
+.EN
+.PP
+the details of this are left as an exercise to the reader, and don't really matter for this
+proof. The only things that matter are that the exponent for b gets larger in the terms not
+listed. If you want to be rigorous, you can try proving this by induction.
+
+.PP
+If you substitute this for the binomial for the derivative definition:
+.EQ
+(x sup {n})' = {x sup n + n{x} sup {n - 1}h + ... h sup n - x sup n} over h
+.EN
+if we cancel out the $x sup n$ terms:
+.EQ
+(x sup {n})' = {n{x} sup {n - 1}h + ... {h} sup {n}} over h
+.EN
+.EQ
+(x sup {n})' = n{x} sup {n - 1} + {... {h} sup {n}} over h
+.EN
+.EQ
+(x sup {n})' = n{x} sup {n - 1} + ... {h} sup {n - 1}
+.EN
+Now we recall that according to our binomial expansion, the exponent
+for $h$ will always grow as we continue looking to the right, and the term $n{x} sup {n - 1} h$ had an exponent of one, which means that each h in the ... will have an exponent of two or more,
+so when we cancel everything out, everything in that ... will
+still have an h term. Because h is infinitely small, we may assume that everything not expanded
+in ... will be almost zero, so our answer here is:
+.EQ
+(x sup {n})' = n{x} sup {n - 1}.
+.EN
+
+.NH 1
+Conclusion
+
+.PP
+Combining knowledge from all of these sections, you will be able to take the derivative
+of an arbitrary polynomial. Next time we will talk about the chain rule and its importance.