From 2459b947d0dbc70ea85acd502688f7ecb1e36bb5 Mon Sep 17 00:00:00 2001 From: preston Date: Tue, 28 Feb 2023 22:32:42 -0800 Subject: add another derivative article --- .../mathematics/calculus/derivative_identities.pdf | Bin 0 -> 31220 bytes .../calculus/source/derivative_identities.ms | 170 +++++++++++++++++++++ 2 files changed, 170 insertions(+) create mode 100644 build/website/mathematics/calculus/derivative_identities.pdf create mode 100644 build/website/mathematics/calculus/source/derivative_identities.ms (limited to 'build/website/mathematics') diff --git a/build/website/mathematics/calculus/derivative_identities.pdf b/build/website/mathematics/calculus/derivative_identities.pdf new file mode 100644 index 0000000..b63f034 Binary files /dev/null and b/build/website/mathematics/calculus/derivative_identities.pdf differ diff --git a/build/website/mathematics/calculus/source/derivative_identities.ms b/build/website/mathematics/calculus/source/derivative_identities.ms new file mode 100644 index 0000000..704034f --- /dev/null +++ b/build/website/mathematics/calculus/source/derivative_identities.ms @@ -0,0 +1,170 @@ +.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. -- cgit