Bam! Mr Tarrou. Well we just got done finding slopes of curved functions, curved lines, curved graphs
Whatever you what to call them.
At a particular point.
Now, what we want to do is expand that just a little bit. And actually find some derivatives.
We are going to do two examples - the same two functions that we used in my previous video.
But, we are not going to find the slope AT a given point. We are going to find something even better.
We are going to find a derivative, which is going to be a new function. A derivative of the original one.
It will tell us very quickly the slope at ANY given value of x. So, if you want to find the slope, if we have a curve,
and we want to find the slope of three or four points, and not just one. We want to find the derivative.
Then, we can take that derivative, plug in x=2, x=5, x=whatever...and *snap* just like that,
we'll have the slope of the original function.
And not have to go through the whole derivative process for every single point.
So, the derivative is a function in which you can plug in any value of x, which is in the original function's
domain, to find the slope of that original function at x. Now, this is what we just got done doing in my
previous video. Finding the slopes of tangent lines. We are still going to walk through the same process.
See, in the original video, they were giving us specific values of a, at which they wanted us to find slope.
Now, we want an expression that will find us slope at ANY given value of x, and pretty much instantaneously.
So, we don't want to find the slope at a particular value of a. We want a derivative, which is going to tell us
the slope no matter what the x value is, as long as it is in the domain of the function.
And, thus, we can't call it the slope of a tangent line. Nope! We are going to call it a derivative!
Or, f prime of x. So, we are going to be given a function. We are going to run it through what is basically the
difference quotient. We are going to find the limit as h approaches zero. And that will be the derivative of the
original function. That derivative will tell us the slope of the original function at any given value of x.
Very, very cool! (If you are a math nerd like I am!)
Okay! First of our two examples...we are just going to find the derivative of two functions.
First one will be f of x is equal to 2x-squared minus 3x plus 1.
Now, there is going to be some algebra involved. Lot's of steps, so be careful of your signs.
Be careful with your parenthesis. And, when you are done, just before you go to find the limit,
as h approaches zero, a lot of cancelations are going to happen. And hopefully, you get your final answer.
So, hopefully it is correct.
We all get answers, they're just not always right. Hopefully we do in this video...
So, f prime of x, or the derivative of the f is equal to...hmmm okay. Again, careful with the notation.
So, f of x+h...so 2 times something-squared, minus 3 times something, plus one. That's the first term of our
numerator. Minus, again, 2 times, well... it's f of x+h. Let me just go ahead and plug that in.
Wherever there is an x, put in an x+h. And then the second term of the numerator, it's x of x+h
minus just f of x. In other words, just copy this function down again.
So, 2x-squared, minus 3x, plus 1. All over, h.
Okay? So, we have x+h...squared. If you often mess up taking a binomial and squaring it,
write it twice and distribute it together. Make sure get that middle term.
This is going to be...whoops! Forgot the limit!
F prime of x is equal to the limit as h approaches zero.
That's the biggest point of this question.
So, the limit as h approaches zero of...take this and distribute it out.
We have 2 times x-squared plus 2xh plus h-squared -  if you carefully distribute that binomial together
with itself. And then take the negative three and distribute it through. So negative three times x is
negative 3x. Negative 3 times h is negative 3h. Plus one. Now, we are going to take this negative, and
distribute it through the second term of the numerator. If you have not used those parenthesis, when you do
substitution, your problem is going to fall apart. You will have the sign errors all over the place.
Negative 2x-squared. Negative times negative is positive 3x. And negative one.
Alllllll ovvvvver h. Whoo!
Take this two and distribute it. Please make sure that you keep working through the notation.
Keep the limit notation going through. You will get to some point in calculus when the problems become long.
(If you don't already think this is long enough.)
And, the less sloppy you are with your notation, the easier it is to forget what you've done or make careless
mistakes. So 2x-squared, plus 4xh, plus 2h-squared, minus 3x, minus 3h, plus 1, minus squared,
plus 3x minus 1... allll ovvveerrr h.
Now at this point, we still have h in the denominator. We need to let h approach zero.
This has to cancel out. It will cancel out through this process. If it doesn't, you've made a mistake.
Now at this point, I've done lots of expansion. Now, we are going to start canceling out.
You need to cancel enough out to...If you do all your cancelation, combine all your like terms.
And you get to a point where you cannot factor the numerator to get the h in the denominator to cancel,
you need to go back through your work to see where you made a mistake.
You should get to a point where that denominator, after factoring, can cancel.
2x-squared and negative 2x-squared cancel.
What else do we get out of it?
Positive 3x and negative 3x. We've got positive one and negative one.
And, you see all my terms that are left have an h.
That's a good sign!
So, we have the limit, as h approaches zero of...2h-squared....plus 4xh...minus 3h.
All over h.
That, with these three terms in the numerator all having a value of h in them, becomes...
The limit as h approaches zero. Take out the h. And you get 2h-squared, divided by h is 2h.
4xh divided by h is 4x. And 3h divided by h is minus 3.
All over h.
Now, look! See? I told you it would cancel out!
The h in the denominator cancels out! So, I can just take the value of zero, plug it into this expression.
And we get...the limit has h approaches zero of 2h plus 4x minus 3 is...
2 times zero, plus 4x, minus 3.
Of course, this is zero. So this is the expression that is left. This is my derivative.
F-prime of x is equal to 4x minus 3.
That is my derivative. Now, what I can do with this, if I want to find my slope of this parabola, at an x value
of zero, just plug in zero! And it would be negative 3.
If I want to find the slope of this parabola at x=2, plug in 2. 4 times 2 is 8, minus 3 is 5.
And, I just found the slope along two points of that parabola.
*Snaps* Pretty much right off the top of my head, using the derivative.
One more example! BAM! This stuff is awesome!
Last example!
Function has a square root symbol in it. So, i'm probably going to have to rationalize the numerator, as I find
the derivative. So, find the derivative of f of x equals the square root of x minus 1.
Well, the derivative of a function is equal to...f of x+h. For the limit as h approaches zero.
I won't forget that again.
Of f of x+h. So, the square root of something, minus one. Minus well, just f of x.
So just copy the function down.
And again, the f of x+h is right there. Take out the x, plug in x+h.
Over h.
Okay. Well, just got a lot of arithmetic, or algebra to do, to get the numerator rationalized.
And hopefully, get to some point where I can get rid of that h in the denominator.
So, we are going to multiply top and bottom by the conjugate.
We just change the little sign in the binomial.
So top and bottom by the square root of x+h minus 1 PLUS the square root of x minus 1.
Okay, distribute.
Doing this very, very slowly and carefully.
Make sure I don't make any mistakes. Hopefully you are doing the same thing.
Square root of x+h minus 1 times the square root of x+h minus 1... when you multiply square roots
they cancel each other out. So, we just get x+h minus 1. I am going to put parenthesis around this to highlight
that was the first term. Now, I've got this square root times this square root, which is positive.
And I've got the exact same square root here, times this same square root, and i've got a minus sign.
This is a difference of squares pattern: a-b times a+b. When you multiply that pattern together,
The middle term cancels out. So, I am going to skip that step, just because of how much writing I am going
to do to finish this. So we have square root of x+h minus 1, times the square root of x+h minus 1, equals...
and then the positive and the negative terms cancel out in the middle.
And then we get negative times a positive, so negative. Parenthesis - square root of x minus one,
times square root x minus one. Again, the square roots are going to cancel each other out.
Because you are multiplying the exact same thing together.
Let's see here...square root of three times square root of three is square root of 9...square root of 9 is still three.
Okay? So they do cancel each other out.
So, minus x minus one. (That binomial is inside the square root's function.
Over, h times all of this.
So.. the h times the square root of x+h minus 1, plus the square root of x minus 1.
Distribute this negative sign.
And cancel some stuff out now.
We have the limit as h approaches zero of...
x minus x. That's going to cancel out.
Negative one, minus negative one is negative one plus one...which cancels out.
And we have h over h times square root of x+h minus 1 , plus the square root of x minus 1.
The h's cancel out. Because it is just h times all of this.
And we get one in the numerator.
Well, now I am going to take this and actually find the limit has h approaches zero.
I can take zero and plug it in here now, and not get a denominator which is zero.
Thus, undefined.
So we get one over the square root of x plus zero (because we are letting h approach zero) minus 1.
Plus, the square root of x minus 1.
Now, zero minus one is negative one.
Now we can actually put these square roots together, because they are EXACTLY the same on the inside.
So we just get one over two times the square root of x minus one.
So, the derivative is f prime of x, is equal to one over two times the square root of x minus 1.
And with all the algebra involved here, hopefully your teacher is not going to make you rationalize the square
root out of the denominator - and are happy that you got the derivative and can take any value of x now, that
you know, for which this is defined and the original function is defined.
Plug any value of x *snaps* boom! You've got the slope of this original function.
I'm Mr. Tarrou! BAM! GO do your homework!
And thank you for watching!
