Differentiation
Remember the dy/dx formula?
A lot of schools use that as just "difference in y". Technically, it is.However, for mathematicians and professionals, the "dy" means an infinitely small change in y. (thus so does dx)
This is going to be needed when the slope of a graph (like that of y=x2) is constantly changing at any given value of x.
I'll show you why that is below.
What does that even mean?
Let me show you.Look at this graph of y=x4 (or just f(x) = x4) where the line is curved and, thus, the slope of it is changing:
Just look for yourself: at X=0.1, the line is pretty much straight, thus the slope is pretty much zero. At X=0.8, the line is very steep (almost fully vertical) and thus the slope is bigger.
If you already know how to differentiate, you'll see that the slope of the line at any given value of X will be:
Slope = 4 × X3
So how do we differentiate?
It's just that: We pick a point along the curve and get the slope of the tangent line (the tangent line here is that red line)
We have to do that because if we try to do "rise over run" across different points in that line, then it's not going to work because we're basically just going to get the average between those two points. This is because the slope at y1 and y2 are different.
But how do we find the slope of a damn point?
Well, we use the idea I talked about before: an infinitely small change in y.
To understand this, first look at this typical "rise over run" mthod of finding the slope:
Notice how on the left it is at point x, then on the right it's (x+h)?
What if we it (x+10)? then the slopes are VERY different, as it is much steeper and thus larger at x+10 compared to x.
Well, what if it's x and x+1?
In this case, the slopes are pretty close together. If we did rise over run here, then we'd get something close to the slope.
So what do we do to get it at 100% accuracy? The smaller the change in x, the more accurate.
How about we make it an infinitely small change?
in that case, we'd have to say that "h" is very close to zero, something like:
0.00000001
It is the same story on the Y axis
Let's say f(x) = x2
For x=2:
f(2) = (2)2 = 4.0
Let's try for f(x+h) at different values of h,
f(2 + 0.5) = (2+ 0.5)2 = 6.25
f(2+ 0.1) = (2.1)2 = 4.41
f(2+ 0.0001) = (2.0001)2 = 4.00040
See, it gets closer and close to 4.0
Try think of this for yourself first in order to understand how it works.
What we just did by looking at "what happens to (x+h)2 as h gets smaller and smaller?" is called limits in calculus.
First think: What'd be the smallest possible value it can take?
10.5
0.1
0.0001
...
It get's closer and closer to what value?
That's right: zero, 0
So we can phrase our question as "What answer are we going to get for (x+h)2 gets closer and closer to 0?"
In other words, we are taking the limit of that equation as h approaches 0
The equation is written as
The answer is just: x2
Congratulations, you now understand the concept of limits in calculus!
Try taking the limit of the following equations before moving on to the next and final step:
Lim X--> 0
5+X -------- 5 + 0 =5
X/1 ------- 0 / 1 = 0
1/X ----------Note that as get's smaller and smaller, the answer get's bigger and bigger.
So, Infinity and undefined as infinity is not a number!
It's that simple.
Let's look back at this for a moment:
First of all, what is it going to look like if we do the rise/run?
Our "y2" is f(x+h) and thus
y1 = f(x)
x1 = x
x2 = (x+h)
To make it accurate, we have to make 'h' as small as we can. So, let's take the limit as it approaches 0:
Let's try this out.
Let's say f(x) = x2
so: (x+h)2 - x2 =
x+h-x = h = 0
Take the limit
0/0 = 1 and the slope clearly isn't always 1!
(You know how 5/5=1 or any number, x/x=1, well, it is the same for all numbers, including any number extremely close to 0)
Now let's say we have that function
f(x) = x2
so f(x+h) = (x+h)2 as we saw above. Let's do the algebra:
Note: when we find the derivative of f(x), we write it as f'(x), as in the photo.
To have a clear understanding of every step of this algebraic working, CLICK HERE----------
where I'll get you to see the logic to it all!
Anyway, make sure you at least try this entire working by yourself.
When we end up with (2x + h), we have the same formula we started off with, but written in a different way.
Now that we have that, we take the limit as you learned:
2x + 0 = 2x
So, therefore, is the slope at any point for the graph of y=x2
eg, at x = 2, the slope is 22 = 4
at x=10, the slope is 102 = 100
What we just did it take literally infinitely small changes in the X and Y to find the slope of an infinitely small point. This has the same effect as taking as finding the slope of the tangent line along any point on the graph!
A shorter way of differentiating
Although that will help to understand what we're dealing with, it's too long.
When we have a formula like
x2 + 5 + 6x + x4
or just as simple as
x6
how the heck will we do that? that would take way too long. So, unless you like to do everything from first principles, mathematicians came up with a neat little rule to differentiate any formula.
This is called taking the derivative, by the way.
Let's take the derivative of f(x) = x2
X is to the power of 2
First, we put the power (2) next to X:
2X2
Then we subtract "1" from the power:
2X(2-1)
=2X1
=2X
We came to the same answer as above: 2X
but much faster.
Look at this from wikiHow:
Questions
Differentiate the following
xn
ANSWER-------- n × X(n-1)
3x5
--------5 × 3 × X(5-1)
= 15X4
X
------= 1 × X1 = X0 = X/X = 1
Look at the last example: It's for f(x) = x or just y=x
What's that slope of that gonna be? For 1 of X, we move by 1 in Y.
Therefore, the slope is always just 1
Differentiating a single number
Let's start by looking at this graph again:
What is the slope? That is, how much does Y go up for each step in the X?
that's right, 0.
We can do the rise over run,
0 - 0 / x = 0
and differentiate too! But how?
Well, since X0 = X/X = 1,
and 1×4 = 4
then
4 = X0 × 4
Let's differentiate that:
(X - 1)×0×4 = 0
So, if you differentiate any number (without X), it's always going to be 0
9 --> 0
10 --> 0
233 --> 0
Differentiating an equation with multiple variables
let's say we have
f(x) = 4x3 + 5 - x
You simply differentiate every variable separately!
4x3 --> 3×4x2 = 12x2
5 = 5 × X0 --> 0×5 × X-1 = 0
-x = -x1 --> -1× X0 = -1×1 = -1
So, that formula differentiated is:
12x2 + 0 - 1
= 12x2 -1
That is then the slope at any given point along the graph of Y = 4x3 + 5 - x
Questions
If we have the equation:
Y = 3X2 - 2X + 4
What'd be the slope at X= -2 ?
Since Y decreases as X increases, then we have a negative Y value for each step in the X for the left side of the curve.
Anyway, if we differentiate it:
dy / dx = 6x - 2
since we're looking for when x = -2,6(-2) - 2 = -12 -2 = -14
Therefore, the slope of the tangent to the point at x=-2 is -14
By the way, we're always differentiation whatever is in the Y axis in respect to whatever is in on X axis.
If we had "stones" in the Y axis, let's call it "S", and meters in the X axis, let's call it "M", then we're looking for dS/dM
The equation of the tangent
we have y2-y1
for (1,2)
etc
lalaal=0
for equation LALAAL, find the point at which the slope is 5
>differentiate it
>answer = 5, solve for X
>y=lalala × X
(x,y) :)
we have y2-y1
for (1,2)
etc
lalaal=0
for equation LALAAL, find the point at which the slope is 5
>differentiate it
>answer = 5, solve for X
>y=lalala × X
(x,y) :)
-Maxiumum and minimum