Topic 2 5 2 6

Ajouté : 2026-05-11

[00:00:05]okay welcome back uh today we're going to go over um some rules that you're going to be using throughout the entire year um for differentiation um first of all we're gonna start with uh the constant rule and you'll hear me say over and over again that the derivative of a constant is always equal to zero the derivative of a constant is always equal to zero let's just quickly talk about why um if you take the graph of any y equals any constant i'm just going to say 5 the graph of y equals 5 or y equals any constant is some horizontal line and what does the derivative represent the derivative represents the slope at any point on that line what's the slope of a horizontal line slope of all horizontal lines is equal to zero so that's why the derivative of a constant is always equal to zero so understanding y makes it much easier to remember them instead of just memorizing them and then forgetting okay then we have the [Applause] derivative of x well if again if it's the line y equals x what is the slope of that line at any point it's always going to be the same it's going to be 1. so the derivative of x is equal to 1. well we can apply the power rule to any kind of polynomial function so x raised to anything not just x raised to the first the power rule is the derivative of x raised to any power is the exponent times x to the n minus one exponent so basically you'll have to remember the power rule is multiply and subtract one from the exponent okay and that's what we are going to do in these next couple examples so pretty straightforward uh the derivative of x or what we refer to as d y d x or f prime of x multiply by the exponents so you bring the exponent out in front and subtract one from the exponent so your derivative is 5 times x to the fourth okay example b well this can be written in rational exponent form hopefully you remember that the root is the denominator and the exponent the power here is the numerator so that's the same thing as the fourth root of x cubed so what is g prime of x that's 3 4 times x and you're going to subtract 1 from the exponent 3 4 minus 4 4 gives you a negative one-fourth okay or we prefer to write that as a positive exponent so that would be x to the positive 1 4 in the denominator if you wanted to change it back and again if it's a multiple choice problem you never know which way they're going to write the answer you need to know that that is the fourth root of x three different ways you could possibly write it and honestly on a free response question they'll accept any of them but on a multiple choice you'll have to make yours look like theirs okay all right so then we have this this last one right here this needs to be rewritten using our algebra 1 rules negative exponent properties that's x to the negative three so the derivative of y you could say y prime uh you could say d y d x we're taking the derivative with respect to x so y prime it's understood that you're taking the derivative with respect to x or again you can write d y d x either notation so multiply and then subtract one when you subtract one you don't get negative two you get negative four so that could also be written as a negative three over x to the positive four so that is uh the power rule okay right moving on to uh a constant what happens when they throw constants in front other than a one well we don't have to worry about that because the derivative of a constant time of func times a function is just the constant times the derivative of that function so again this 2 right here we can kind of ignore and the derivative of y is y prime or d y d x so the constant times the derivative of this right here so that's 7 x to the sixth or an answer of 14 x to the sixth okay so this needs to be for now this needs to be rewritten as a negative exponent and we can take the derivative of this function right here and multiply it by three so the derivative of g of x is g prime of x and we're just going to take our constant and we're going to multiply it by the derivative of x to the negative 2.

[00:06:39]so that's multiply by the exponent and subtract 1. that gives you a negative 3 exponent okay so that's a negative 6 x to the negative 3 or a negative 6 over x to the positive 3. negative exponent properties are not negative coefficient properties so please do not take the negative six and move it to the denominator you are telling me that a negative six is equal to 1 over 6.

[00:07:16]so do not make that algebra mistake all right then we have this one this one can be written as 1 8 times x to the what exponent 5 6 so the derivative of f is f prime of x and we have the 1 8 constant we're going to take the derivative by multiplying and subtracting one 5 6 minus 6 6 is a negative 1 6.

[00:07:58]okay and that will be written as 5 over 48 um x to the positive 1 6 or the sixth root of x okay so that is if you have a constant in the front it's not that big of a deal all right so then we're going to get into um sum and difference okay the sum and difference are are pretty basic again if you are adding two functions together you can take the derivative of the first and the derivative of the second and add them together so that's the sum rule same thing with the difference rule okay now this right here is a quotient that means division there's something you're going to be learning called the quotient rule and you want to avoid the quotient rule and you want to avoid the product rule whenever possible because those take longer so i'll say that over and over again avoid the quotient rule and avoid the product rule what i want you to understand is that you cannot take the derivative of the top and the derivative of the bottom and divide them that's not going to work you can't take the derivative of the first and the derivative of the second and multiply them you will learn that in the next couple of topics so in order to be able to do these problems and when you do know the product and quotient rule you always want to see if you can avoid them at all costs so is there a way to make this not a quotient you always want to see if you can simplify first simplify first whenever possible before taking derivatives so if you remember anything from algebra this is a common denominator for each one of those terms so we can actually split this up as x cubed divided by x minus 4x divided by x plus 5 divided by x okay so our f of x is x squared minus 4 plus five i'm going to rewrite five over x and make it a little bit easier that's going to be five instead of x to the first on the bottom that's x to the negative one in the numerator so instead of x in the denominator i moved it to the numerator using my algebra one properties if you need to review negative exponent properties you can look that up on youtube or khan academy okay so i simplified this problem to make it easier to do you always want to try to simplify first so now when taking the derivative f prime of x you're going to multiply and then subtract one which would give you x to the first the derivative of a constant is zero and then the derivative of this one you're going to multiply and that's going to give you negative 5 and then you're going to subtract 1. that's the power rule that's going to give you an exponent of negative 2.

[00:11:28]okay which is the same thing as two x minus five over x to the positive two again using your negative exponent properties okay so you can take the derivatives of each term and add them and or subtract them but you cannot take the derivative of the top divided by the derivative of the bottom and same thing here product does not work so we can't just take the derivatives we have to make them a polynomial individual terms so on this one i can use foil and i get x cubed my outside is three x squared my inside is positive x and then negative three so now i have a sum and a difference and i can use my sum and difference derivatives which says i can do each one individually so you'll get that fast again power rule three times x subtract one so multiply subtract one multiply subtract one from the exponent the derivative of x is equal to one that's right here now again you're using the power rule on this one if you want to think about it if you multiply by one and then you subtract one from the exponent what do you get you get x to the zero power and anything raised to the zero power is one so you could apply the power rule to this but it's just easier to understand that the derivative of x is always equal to one and i showed you that earlier and then the derivative of a constant is always zero okay so those are your basic derivative rules that you will need throughout the entire year okay all right moving on to example number four uh writing equations of tangent lines um again we'll be doing an activity um and looking at tangent lines and what all the all of these things mean and put it into applications um so let's review what you learned back in algebra one again um to write the equation of a tangent line or to write the equation of a line there are two things that you will always need and again this is a reoccurring question throughout the entire year write the equation of a tangent line you need two things you need a point and a slope you need a point and a slope well they give you a point but how are you going to find the slope at that point well everything that we've learned up to this point has told you that the derivative represents the slope on the graph for any point so if we take the derivative of f of x the derivative of x is one minus four x again that's the power rule two times two is four subtract 1. so this is the derivative or the slope of the tangent line at any x value this is a formula to find the slope at any point on the graph so they want us to find the slope at x equals one so we're going to evaluate f prime of one which is one minus four times one which gives you a slope of negative three so now we have a point and now we have a slope at that given point so we are going to write the equation of the line now one form of the line you've learned standard form you've learned slope-intercept form and you should have learned point-slope form for those of you who don't remember point slope form go ahead and write it up here because we always use this in calculus because it's easy and it's less work point slope says y minus y1 equals your slope times x minus x1 that is point slope now if you think about it where does this come from that comes from slope being change in y's over change in x's well if a slope is equal to change in y's over change in x's couldn't i just multiply this denominator by both sides and i get this equation right here that is called point slope form and that is what we will use all year so to answer this question what is the equation of the line it's y minus the given y value of negative one so y minus a negative one is y plus one equals slope times x minus the given x value and we are going to leave it we're not going to rearrange it we're not going to change it to slope-intercept form we're just going to leave it save ourselves time and save ourselves from making mistakes okay all right it is a very common mistake in example like and this is supposed to be number four that's a typo there for part a is to think that the slope of the specific tangent line is y is one minus four x so when it says write the equation of the tangent line i will get y equals one minus four x okay that is not true the derivative represents think of this as a formula to find the slope of the tangent line it is not the equation of the tangent line okay and that's what this is just restating right here so don't make that common mistake all right so you can try the next one if you want to if you want to pause the video and give yourself a chance to try that and i will do it with you okay so it says write the equation of a tangent line to the function at the given value of x okay as i said earlier you need a point and you need a slope okay they give us an x value we need a y value to go with it and we need a slope okay so if i give you an x value don't overthink it okay if i give you an x value this means y this is a fancy way of saying y so if i plug 1 into here i get 2 times the square root of 1 it gives me a 2. there's my point so that's basic algebra 1. given an x value find the y value okay now we need a little bit of calculus find the slope at this point okay again the derivative represents the slope of a tangent line so let's go ahead and find the derivative of two square roots of x two square roots of x well the only thing we know is the power rule right now so you're going to have to rewrite this i'm going to give you a shortcut for this later but for right now you have to rewrite this as x to the one half power again you need to know it a radical and a rational in how they're related way back to algebra one again so now we take the derivative and we multiply and we subtract one so that tells us that our derivative is x to the negative one-half or one over x to the positive one-half or one over the square root of x that is the derivative the first derivative is the formula to find the slope of the tangent line at any x value okay so we need to find f prime of one because that's what we want to evaluate f prime of one is one over the square root of one or just one okay so go ahead and write the equation of the tangent line at one y minus the y value of your point equals your slope times x minus the x value of the point and of course you do not have to write the one in the front okay you do not have to write the one in the front i just wanted to put it in there just to remind you that that's where the slope goes so you can leave it like that or you can write y minus two equals x minus one and that's the slope or i'm sorry that's the equation of the tangent line to that graph right there so let's just take a quick look at that i'm gonna i'm gonna go here um let's go ahead and and you can even try it on your calculator if you want to the square root function one usually one one but we're multiplying it by two if we go two three four it's usually four two but we're multiplying it by 2.

[00:23:09]okay so there is f of x that was given to us that's the sketch of f x okay and what you are finding is the equation of this line right here that's what this equals now it's very easy for us to solve this for y you don't have to but i'm just going to add 3 i'm sorry add 2 to both sides and that's going to give me what we call slope intercept form which again i would never change to slope intercept form because on the ap exam you're not going to have time and it just if you make a mistake by switching it around they're not going to give you credit so we just usually leave it like this but if i add 2 to both sides i end up getting y equals x plus 1.

[00:24:10]that's what you just found you found well look at this there's your y intercept of one and up one over one has a slope of one that's what you just found the equation of that tangent line to the curve at x equals one you have an idea of what you're actually doing okay all right last example at what point or points does this graph have a horizontal tangent line horizontal tangent line well what do you remember from algebra about horizontal lines well i said it earlier horizontal tangent lines have a slope equal to zero so they want to know where is this going to be equal to zero okay well let's go ahead and do find the slope formula for this equation well again what does that mean that means do the calculus so y prime the derivative of y is two x plus four again power rule derivative of x is just equal to one times four and derivative of a constant is equal to zero so we had a power rule we had the derivative of x and we have the derivative of a constant all right so this is the equation for the first derivative which represents the slope of the tangent line at any x well they're asking us when is that going to be equal to zero well if we subtract 4 and we whoops positive 2 if we subtract 4 and then divide both sides by 2 we get x is equal to a negative 2. it says at what points it doesn't say at what x value it says at what points well we know when the x value is equal to negative 2 the slope or the first derivative the slope of the tangent line will be equal to zero so how do we find the y value again well okay this is the equation to find the y value that goes with it you plug in the x value just common sense basic algebra if you know the x value plug it into the function to find the y value and that's going to give you 4 minus 8 minus 1 which is a negative 5.

[00:26:58]so the answer is at one point okay at what point or points does this graph have a horizontal tangent line and the point is negative two negative five what kind of graph is this that's a quadratic equation what kind of graph do you get you get a parabola okay so in advanced algebra two you should have learned how to graph parabolas and find the vertex um completing the square opposite of v over 2a which is what most of you remember so let's go back to algebra if i gave you the equation y equals x squared plus four x minus one most people did not want to complete the square to change it to vertex form which is fine so let's go the x value of your vertex was equal to the opposite of b over two times your a value hopefully that sounds familiar to you so the opposite of four over two times your a value our a value is a one gives us negative 2.

[00:28:45]and look at that our algebra gave us the same answer as our calculus and then if you plug the negative 2 back into there you get the vertex of negative negative 2 negative 5.

[00:29:04]so what we have here in essence is negative 2 1 2 3 4 5. negative 2 5 is the vertex and this is a parabola that opens up when x is equal to zero circles here here if you put on your graphing calculator if you want to but you learned how to do that in algebra what we just now did in calculus is we found the point which in algebra you called the vertex we found the vertex of the parabola using calculus okay well we don't just deal with parabolas we deal with all kinds of functions in calculus so that's why the calculus method the first derivative is going to help us easily find the point where your graphs have a horizontal tangent line okay all right and that is the end of these notes