Now that we've seen what a derivative means, and what it has to do with rates of change.
Our next step is to learn how to actually compute these guys, as in if I give you some kind of function
with an explicit formula you'd want to be able to find what the formula for its derivative is.
Maybe it's obvious, but I think it's worth stating explicitly why this is an important thing
to be able to do. Why much of a calculus students time ends up going towards grappling with
derivatives of abstract functions rather than thinking about concrete rate of change problems,
Is because a lot of real-world phenomena. The sort of things that we want to use calculus
to analyze are modeled using polynomials, trigonometric functions, exponential's
and other pure functions like that. So if you build up some fluency with the ideas
of rates of change for those kinds of pure abstract functions. It gives you a language to more
readily talk about the rates at which things change in concrete situations that you might be
using calculus to model. But it is way too easy for this process to feel like just memorizing a list
of rules. And if that happens, if you get that feeling it's all so easy to lose sight of the fact that
derivatives are fundamentally about just looking at tiny changes to some quantity,
and how that relates to a resulting tiny change in another quantity. So in this video
and in the next one, my aim is to show you how you can think about a few of these rules intuitively
and geometrically. And I really want to encourage you to never forget that tiny nudges
are at the heart of derivatives. let's start with a simple function like f(x) = x^2, what if I asked
you it's derivative. That is if you were to look at some value x like x = 2 and compare it to a value
slightly bigger, just dx bigger. What's the corresponding change in the value of the
function df, and in particular what's df divided by dx - The rate at which this function is changing
per unit change in x as a first step for intuition. We know that you can think of this ratio df/dx
as the slope of a tangent line to the graph of x^2, and from that you can see that the slope
generally increases as x increases. At 0 the tangent line is flat so the slope is 0 at x = 1.
That's something a bit steeper at x = 2, it's steeper still but looking at graphs isn't generally the
best way to understand the precise formula for a derivative, for that it's best to take
a more literal look at what x^2 actually means. And in this case let's go ahead and picture a square
whose side length is x if you increase x by some tiny nudge, some little dx.
What's the resulting change in the area of that square, that slight change, in area
is what df means in this context. It's the tiny increase to the value of f(x) = x^2 caused by
increasing x by that tiny nudge dx. Now you can see that there's three new bits of area in this diagram, two thin
rectangles and a miniscule square. The two thin rectangles each have side lengths of x and dx
so they account for two times x * dx units of new area. For example let's say x was 3 and dx was 0.01.
Then that new area from these two thin rectangles would be 2 * 3 * 0.01 which is 0.06,
about 6 times the size of dx. That little square there has an area of dx^2, but you should
think of that as being really tiny, negligibly tiny. For example, if dx was 0.01, that would be only 0.0001.
And keep in mind, I'm drawing DX with a fair bit of width here just so we can actually
see it, but always remember - in principle dx should be thought of as a truly tiny amount
and for those truly tiny amounts. A good rule of thumb is that you can ignore anything that includes a dx
raised to a power greater than 1. That is a tiny change squared is a negligible change what this leaves us with
is that df is just some multiple of dx, and that multiple 2x which you could also write as df/dx
is the derivative of x^2. For example, if you were starting at x = 3, then as you slightly increase x
the rate of change in the area per unit change in length added dx^2/dx would be 2 * 3 or 6.
And if instead you were starting at x = 5, the rate of change would be ten units of area per unit change in x.
Let's go ahead and try a different simple function f(x) = x^3. This is going to be the geometric view
of the stuff that I went through algebraically in the last video. What's nice here is that we can think of x^3
as the volume of an actual cube, whose side lengths are x. And when you increase x by a tiny nudge, a tiny dx
the resulting increase in volume is what I have here in yellow. That represents all the volume in a cube
with side lengths x plus dx. That's not already in the original cube, the one with side length x.
It's nice to think of this new volume as broken up into multiple components, but almost all of it comes
from these three square faces, or set a little more precisely as dx approaches zero.
Precisely as dx approaches zero, those three squares comprise a portion closer and closer to 100%
of that new yellow volume. Each of those thin squares has a volume of x^2 * dx,
the area of the face times that little thickness dx. So in total this gives us 3x^2 dx of volume change.
And to be sure there are other slivers of volume here, along the edges, and that tiny one in the corner. But all
of that volume is going to be proportional to dx^2 or dx^3 so we can safely ignore them.
Again this is ultimately because they're going to be divided by dx, and if there's still any dx remaining
then those terms aren't going to survive the process of letting dx approach 0. What this means is
that the derivative of x^3, the rate at which x^3 changes per unit change of x is 3x^2.
What that means in terms of graphical intuition is that the slope of the graph of x^3
at every single point x is exactly 3x^2. And reasoning about that slope, it should make sense
that this derivative is high on the left, and then zero at the origin, and then high again as you move to the right.
But just thinking in terms of the graph would never have landed us on the precise quantity 3x^2.
For that we had to take a much more direct look at what x^3 actually means now in practice.
You wouldn't necessarily think of the square every time you're taking the derivative of x^2. Nor would
you necessarily think of this cube whenever you're taking the derivative of x^3, both of them fall under a pretty
recognizable pattern for polynomial terms the derivative of x^4 turns out to be 4x^3. The derivative
of x^5 is 5x^4. The derivative of x to the n for any power n is nx^(n-1).
This right here is what's known in the business as the power rule. In practice
we all quickly just get jaded and think about this symbolically as the exponent
hopping down in front leaving behind one less than itself. Rarely pausing to think about the
geometric delights that underlie these derivatives. That's the kind of thing
that happens when these tend to fall in the middle of much longer computations.
But rather than tracking it all off to symbolic patterns let's just take a moment and think about why this works.
For powers beyond just 2 and 3 when you nudge that input x, increasing it slightly to x + dx, working out the
exact value of that nudged output would involve multiplying together these n separate x + dx terms.
The full expansion would be really complicated but part of the point of derivatives is that most of that
complication can be ignored. The first term in your expansion is x^n, this is analogous to the
area of the original square or the volume of the original cube from our previous examples.
For the next terms in the expansion, you can choose mostly x's with a single dx since there are
n radicals from which you could have chosen that single dx. This gives us n separate terms all of which include
n-1 x's times a dx giving a value of x^(n-1) times dx. This is analogous to how the majority of
the new area in the square came from those two bars each with area x * dx,
or how the bulk of the new volume in the cube came from those three thin squares.
Each of which had a volume of x^2 times dx. There will be many other terms of this expansion
but all of them are just going to be some multiple of dx^2 so we can safely ignore them.
And what that means is that all but a negligible portion of the increase in the output
comes from n copies of this x to the (n-1) * dx - that's what it means.
For the derivative of x^n to be n * x^(n-1) and even though, like I said in practice, you'll find
yourself performing this derivative quickly and symbolically imagining the exponent,
hopping down to the front. Every now and then it's nice to just step back and remember why these rules work.
Not just because it's pretty, and not just because it helps remind us that math actually makes sense
and isn't just a pile of formulas to memorize. But because it flexes that very important muscle of
thinking about derivatives in terms of tiny nudges. As another example think of the function f(x) = 1/x.
Now, on the one hand you could just blindly try applying the power rule since 1/x
is the same as writing x^(-1). That would involve letting the negative 1 hop down in front
leaving behind 1 less than itself which is -2. But let's have some fun and see if we can reason about this
geometrically rather than just plugging it through some formula. The value 1/x is asking what number
multiplied by x equals 1, so here's how I'd like to visualize it - imagine a little
rectangular puddle of water sitting in two dimensions whose area is 1, and let's say that its width is x.
Which means that the height has to be 1/x since the total area of it is 1, so if x was
stretched out to 2 then that height is forced down to 1/2 and if you increased x up to 3
then the other side has to be squished down to 1/3 this is a nice way to think about the graph of 1/x.
By the way, if you think of this with x of the puddle as being in the xy plane then that corresponding output 1/x,
the height of the graph above that point is whatever the height of your puddle has to be to maintain an area of 1.
So with this visual in mind for the derivative, imagine nudging up that value of x by some tiny amount,
some tiny dx. How must the height of this rectangle change so that the area of the puddle remains
constant at 1 - That is increasing the width by dx, add some new area to the right here so the puddle
has to decrease in height by some d(1/x) so that the area lost off of that top cancels out the area gained.
You should think of that d(1/x) as being a negative amount. By the way since it's decreasing the height
of the rectangle and you know what I'm
going to leave the last few steps here
for you for you to pause and ponder and
work out an ultimate expression and once
you reason out what D of 1 over X / DX
should be I want you to compare it to
what you would have gotten if you had
just blindly applied the power rule
purely symbolically
to
X to the negative one and while I'm
encouraging a deposit ponder here's
another fun challenge if you're feeling
up to it see if you can reason through
what the derivative of the square root
of x should be
to finish things off I want to tackle
one more type of function trigonometric
functions and in particular let's focus
on the sine function so for this section
I'm going to assume that you're already
familiar with how to think about trig
functions using the unit circle the
circle with the radius one centered at
the origin
for a given value of theta like say zero
point eight you imagine yourself walking
around the circle starting from the
rightmost point until you've traversed
that distance of zero point eight in arc
length this is the same thing as saying
that the angle right here is exactly
theta radians
since the circle has a radius of one
then what sine of theta means is the
height of that point above the x-axis
and as your theta value increases and
you walk around the circle your height
Bob's up and down between negative one
and one so when you graph sine of theta
versus theta you get this wave pattern
the quintessential wave pattern
and just from looking at this graph we
can start to get a feel for the shape of
the derivative of the sine the slope at
zero is something positive since sine of
theta is increasing there and as we move
to the right and sine of theta
approaches its peak that slope goes down
to zero then the slope is negative for a
little while while the sine is
decreasing before coming back up to zero
as the sine graph levels out and as you
continue thinking this through and
drawing it out if you're familiar with
the graph of trig functions you might
guess that this derivative graph should
be exactly cosine of theta since all the
peaks and valleys line up perfectly with
where the peaks and valleys for the
cosine function should be and spoiler
alert the derivative is in fact the
cosine of theta but aren't you a little
curious about why it's precisely cosine
of theta I mean you could have all sorts
of functions with peaks and valleys at
the same points that have roughly the
same shape but who knows maybe the
derivative of sine could have turned out
to be some entirely new type of function
that just happens to have a similar
shape well just like the previous
examples a more exact understanding of
the derivative requires looking at what
the function actually represents rather
than looking at the graph of the
function so think back to that walk
around the unit circle having traversed
an arc with length theta and thinking
about sine of theta as the height of
that point
now zoom in to that point on the circle
and consider a slight nudge of D theta
along their circumference a tiny step in
your walk around the unit circle how
much does that tiny step change the sign
of theta how much does this increase D
theta of arc length increase the height
above the x-axis
well zoomed in close enough the circle
basically looks like a straight line in
this neighborhood so let's go ahead and
think of this right triangle where the
hypotenuse of that right triangle
represents the nudge D theta along the
circumference and that left side here
represents the change in height the
resulting D sine of theta now this tiny
triangle is actually similar to this
larger triangle here with the defining
angle theta and whose hypotenuse is the
radius of the circle with length one
specifically this little angle right
here is precisely equal to theta radians
now think about what the derivative of
sine is supposed to mean it's the ratio
between that D sine of theta the tiny
change to the height divided by D theta
the tiny change to the input of the
function and from the picture we can see
that that's the ratio between the length
of the side adjacent to the angle theta
divided by the hypotenuse well let's see
adjacent divided by hypotenuse that's
exactly what the cosine of theta means
that's the definition of the cosine
so this gives us two different really
nice ways of thinking about how the
derivative of sine is cosine one of them
is looking at the graph and getting a
loose feel for the shape of things based
on thinking about the slope of the sine
graph at every single point and the
other is a more precise line of
reasoning looking at the unit circle
itself for those of you that like to
pause and ponder see if you can try a
similar line of reasoning to find what
the derivative of the cosine of theta
should be in the next video I'll talk
about how you can take derivatives of
functions who combine simple functions
like these ones either as sums or
products or function compositions things
like that and similar to this video the
goal is going to be to understand each
one geometrically in a way that makes it
intuitively reasonable and somewhat more
you