WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP calculus.
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Today, we are going to continue our discussion of ways of taking the derivative of different types of functions very quickly.
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We are going to discuss the product rule, the power rule, and the quotient rule for functions.
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Let us jump right on in.
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Here is what we have got, what we can do.
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You know what, I think today I’m going to work in black.
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What do we do when we have to differentiate functions that look like this, differentiate functions like the following.
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Let us say f(x) = 2x² – 1³ × 3x – 4⁻³.
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How do we differentiate something like that?
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Or what if we had f(x) = x² × the sum of x.
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Or what if we had f(x) = 2x² + 3x + 2/ x² - 6.
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How do we differentiate that?
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Or f(x) = 3x² + 1⁵.
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How do we deal with these?
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In the first and second cases, in other words, this one and this one, what we have is our function F(x) is actually a product of 2 separate functions of x.
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First case, f(x) is a product of 2 independent functions of x.
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In other words, let me go to red here.
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We have this is one function, the 2x² – 1³.
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This 3x – 4⁻³ is a separate function.
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In this case, we have x² which is an entirely separate function of x and sin x which is an entirely separate function of x.
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When they are multiplied together, how do we take the derivative of something like that?
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In the third case, we have an independent function of x divided by an independent function of x.
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This one over here, we have some independent function of x and it is actually raised to a power.
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We actually have the same case here.
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This whole thing is an independent function of x.
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It is actually made up of a function of x which is raised to a power times a function of x raised to a power.
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How do we handle things like this?
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You might think to yourself, if f is the product of two functions, if I take the derivative of f, it is not just the product of the derivatives?
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In other words, why cannot I just take the derivative of x² multiplied by the derivative of sin x.
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It turns out that that is actually not how it happens.
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Let me write this down.
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You might think that, if f(x) = f(x) × g(x), then f'(x) = f'(x) × g’(x).
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This is not true, absolutely not true.
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It is true that the derivative of the sum is the sum of the derivatives but the derivative of a product of two functions is not the product of the derivatives.
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It also is not true that the derivative of a quotient of two functions is equal to the quotient of the derivatives.
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The derivative of f/g does not equal f’/g’.
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Here are the rules for handling product of functions, powers of functions, and quotients of functions.
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Let us go ahead and go through these.
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I think I’m going to go back to black here.
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Our product rule is, if f(x) and g(x) are differentiable, our presumption on the entire calculus course is that they are going to be differentiable,
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then the derivative of fg is equal to the derivative of f × g + f × the derivative of g.
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In other words, the derivative of this × this + the derivative of this × this.
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You can keep them in order, essentially what you are doing is you are just taking the derivative of one function at a time, as you go down.
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This is a shorthand notation.
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The derivative of the product is f’ × the other function + f × g’, the derivative of the other function.
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This can be extended to a product of 3 or more functions.
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That is going to look like this.
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Let us go ahead and write.
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Let f1 f2 all the way to fn, whatever n happens to be -- 5, 10, 15, 20, however many functions.
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F sub n, let them be n differential functions.
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Then, the derivative of f1 × f2 × all the way to fn = f1’ × f2 × all the functions + f1 × f2’
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× f3 × all the functions + f1 f2 all the way up to fn-1 fn’.
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Basically, all I'm doing is I'm leaving the functions as is and I’m just taking the derivative of the first function.
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And then, I take the derivative of the second function, multiplied all out.
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The derivative of the 3rd function, multiply all out, until I have taken the derivative of every single individual function.
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Of course, I'm adding them.
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Most of the time, we are just going to be dealing with 2 functions.
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This one right here is sufficient.
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That is the product rule, that is how you handle.
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Take the derivative of the product of 2 functions.
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What you have to do with is recognize which 2 functions you are dealing with.
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What is f and what is g.
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All of this will make perfect sense, once we actually see the examples.
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Let us talk about the quotient rule.
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As we said, the derivative of the quotient of 2 functions is not the quotient of the derivatives.
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The quotient rule is the following.
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Ddx (f/g) is equal to the denominator × the derivative of the numerator - the numerator × the derivative of the denominator/ g².
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F and g are functions, these are actual functions.
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Let us go ahead and finish off with the power rule.
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I’m going to use u instead of f or g.
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Let u be differentiable.
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In other words, u(x) is an actual full function of x.
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Let u(x) be differential.
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The derivative of u ⁺x raised to some power, any real number is equal to,
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It is handled the same way as a function of x.
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It is n × the function, the exponent n -1.
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I’m going to multiply by du dx.
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Let us talk about what this actually looks like.
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Before, we dealt with something like this, we said x⁵.
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When I took the derivative, we ended up with 5x⁴.
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We took this, brought it down, subtracted by 1.
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X is the function, it is raised, it is variable.
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It is raised to the 5th power.
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What we are saying is, if you have something like this.
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If you have 2x² + 1⁵.
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This is your u(x).
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It is an entire function that contains x, itself is raised to a power.
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When you take the derivative of this, you treat it the same way.
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It is still going to be 5 × 2x² + 1⁴, except you have to multiply by the function itself.
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By the derivative of the function itself, du dx.
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In this particular case, the du dx, the derivative of what is inside is, that is fine, du dx.
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This ends up equaling 5 × 2x² + 1⁴.
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The derivative of that is 4x.
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It is just that extra step.
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When the thing that you actually raising to a power is a full function of x,
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you have to take the derivative of that function.
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It is the same thing here.
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What we are doing here, it is actually a special case of this x⁵.
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When you take the derivative of this, what you are doing is you are doing 5 x⁴.
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This is your function of x, you are taking dx of dx.
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Dx/dx is just 1.
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It stays 5x⁴.
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You are still doing the same thing except your function of x just happens to be x alone.
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It is the variable itself.
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There is nothing else going on there.
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In any case, I would not be able to point here.
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It will make a lot more sense when we actually do the example problems.
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All you have to remember is once you identify some functions that is being raised to a power,
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you do everything the same way to do the power part.
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And then, you just remember to multiply by the derivative of the original function itself.
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Let us go ahead and get started with our examples.
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Y = x² + 4 × f⁴ + 8, find dy dx.
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There is a couple ways that we can do this that is nice.
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Now, you have multiple paths that you can follow.
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Do I do the product rule, do I multiply it out and just do the normal,
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the derivative of each individual term because it is a polynomial.
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If you want, you can just multiply this.
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This × that, this × that, this × that, and then take the derivative.
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Or you can treat it like a product rule.
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Let us go ahead and do it as a product rule.
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This is our f and this is our g.
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The product rule says that y’ is equal to f’ × g + f × g’.
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Let us go ahead and do that.
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Y’ = f’ × g.
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F’, the derivative of this is 2x × g.
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G is x⁴ + 8, we leave g alone.
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We leave f alone.
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We write x² + 4 and we do g’.
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This is 4x³.
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That is it, you can just leave it like this, if you want to.
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It is perfectly valid.
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Or again, depending on what it is that your teacher or professor wants.
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You can multiply it out.
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If you want to multiply it out, this is just going to be y’ = 2x⁵ + 16x + 4x⁵ + 16x³.
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2x⁵ + 4x⁵, y’ = 6x⁵ + 16x³.
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I will do it in descending powers of x, + 16x.
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There you go, that is it, nice and simple, product rule.
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F’ × g + f × g’.
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All you have to do is recognize what the f and g are.
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That is going to be your ultimate task.
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Y = x +1/ x-1.
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This one, we have f here, we have g here.
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I will go ahead and do this in red.
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This is going to be quotient rule.
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The quotient rule says that y’ is equal to, this is f, this is g.
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G f’-f g’/g², denominator × the derivative of the numerator - the numerator × the derivative of the denominator/ the denominator².
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Let us do it.
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Y = g which is x-1 × the derivative of the numerator which is 1 – f,
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which is x + 1 × the derivative of the denominator which is 1, the derivative of x – 1 is 1, all over the denominator², x-1².
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If I distribute the negative sign, i on top, I get x -1 - x-1/ x-1².
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X cancels, I’m left with -2/x-1².
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That is my y’, that is my dy dx.
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At a given point, x, if I choose a point and I put it in here, this is going to be the slope of my tangent line to that curve.
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This represents the rate of change.
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For a unit change, for a small change, I’m making xy, at that point changes by that much.
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Rate of change, slope.
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Let us try this one.
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We have this thing raised to the 5th power.
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In this particular case, this is going to be our u(x) and it is a whole function raised to the 5th power.
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We are going to use our power rule.
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Our power rule says, f⁵ = 5 × f⁴ × f’.
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Let us go ahead and do it.
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Y’ =, that comes down, 5 ×, I will leave that alone, 4x³ – 3x² +6x + 9⁴ × the derivative × f’.
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Now I take the derivative of what is inside, × 12x² - 6x + 6.
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I’m done, I’m not going to simplify this anymore.
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I’m not going to multiply it out, that is it.
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I’m done, I just have to run through the rule.
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Nice and straightforward.
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Example number 4, y = 5/3x – 5⁴.
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It is a quotient, in the sense that you have 5 over that.
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You can certainly use the quotient rule.
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Another way to handle this is actually doing just the power rule, believe it or not.
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Because I can actually write this as 5 × 3x – 5⁻⁴.
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You have multiple paths that you can take.
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You can do quotient rule or you can do power rule.
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There is always going to be the case like this, with problems like this.
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You are going to have a choice and the choice does not matter.
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Sometimes it is the same amount of effort.
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No matter if you go this path or this path.
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Sometimes, one of them is going to be a lot quicker than the other.
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Sometimes one of them is going to be messy, the other one is going to be not so messy.
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It just depends and this is a question of experience as to which one is going to be better.
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Oftentimes, you do not know which one is going to be better.
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You just have to jump in, go down the path.
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If it looks to be too complicated, you change path and you go down another path, very simple.
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Let us go ahead and do y’ here.
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Y’ = 5, this is a constant.
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This is going to be, I’m going to bring this down.
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This is going to be -4 × 3x-5.
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I’m going to take the derivative of this to be -4, -4 -1⁻⁵ × the derivative of this which is 3.
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Let me do it over here because I need some room. 5 × -4 is -16.
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-16 × 3, 5 × -4 is -20.
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-20 × 3 is -60.
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Let us not jump to going here.
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-60 × 3x -5⁵, this is our answer.
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Let us also do this via the quotient rule to see what it looks like.
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Let us also do this via the quotient rule.
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Remember, we said what the quotient rule is.
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If this is f, this is g, it is g f’- f g’/ g².
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Y’ equals this, 3x -5⁴ × the derivative of that f’ which is 0, -5 × g’ the denominator,
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× the derivative of this which is the derivative of the denominator is 4 × 3x – 5³.
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4-1 is 3 × the derivative of what is inside which is 3/ this², 3x – 5⁸.
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This is equal to, this is just 0.
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We have -60 to be 3x -5³/3x – 5⁸.
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That leaves 5, you are left with y’ = -60/3x-5⁵.
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There you go, this is the same as that.
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It is just expressed with a negative exponent.
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The only thing that you have to watch out for is watch this exponent very carefully.
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If it is negative that -1 becomes more negative.
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Notice, we went from -4 to -5.
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Here, we left it as is, the denominator.
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When we took this g‘, f4 came here.
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4 -1 is 3, you just have to watch the signs.
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If it is positive, it goes down.
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If it is negative, it still goes down, becomes further negative.
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That is the only thing you have to watch out for.
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Again, you are going to make a mistake.
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I made a mistake, I still make a mistake after all these years.
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It is just the nature of the game.
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We just have to be as vigilant as possible.
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Let us go ahead and go on here.
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Let us see, now we have y = this thing.
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It looks like we do have just to do the quotient rule.
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We will just do quotient rule.
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Once again, recall quotient rule.
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Let me go back to red here.
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Quotient rule, y’, if this is f and this is g, we have g f’- f g’/ g².
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Y’ or dy dx is equal to this × the derivative of that.
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We have x³ – 3x – 2 × the derivative of this which is 2x – this x² -5 × the derivative of this
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which is 3x²/ x³ -3x – 2².
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We have to multiply all of this out on top.
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Y’ = 2x and x³ that should be 2x⁴ – 6x² – 4x, this minus sign, x².
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3x² is 3x⁴, the minus sign becomes -3x⁴ x² -3.
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This is -3x², - and – is +, 3x².
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This × this is -15x², it is going to be a +15x².
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This is -5 × -3 is +15.
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It is -15/ x³ – 3x – 2².
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When we put it together, 2x⁴- 3x⁴, that takes care of these.
00:27:20.600 --> 00:27:32.900
We get a –x⁴ -6x² + 3x² is -3x² + 15x⁵.
00:27:32.900 --> 00:27:40.300
We get a +12x² and then we have -4x.
00:27:40.300 --> 00:27:49.400
And then, we have our -15/ x³ - 3x -2.
00:27:49.400 --> 00:27:56.600
That is y’ or dy dx.
00:27:56.600 --> 00:27:57.400
There you go.
00:27:57.400 --> 00:28:06.000
Whenever you are dealing with quotient rule, as we get them to more complex functions, trigonometric functions, things like that, exponentials,
00:28:06.000 --> 00:28:13.400
you are going to notice that the quotient rule, the biggest problem is going to be where do I stop simplifying.
00:28:13.400 --> 00:28:16.000
Again, it is up to your teacher what they want.
00:28:16.000 --> 00:28:19.700
Sometimes, they will just ask you to stop right there and just leave it like that.
00:28:19.700 --> 00:28:25.400
Other times, they will ask you to simplify as much as possible, personal choice.
00:28:25.400 --> 00:28:39.700
The mathematics itself, the important part is being able to recognize the f and g, and to keep the order straight, that is what matters.
00:28:39.700 --> 00:28:48.000
Suppose that u and x and v(x) are functions that are differentiable at x = 0.
00:28:48.000 --> 00:28:52.500
I think I will actually work in blue here.
00:28:52.500 --> 00:29:00.000
U(0) = 7, U’ (0) = -2, v(0) =-3, v’(0) = 2.
00:29:00.000 --> 00:29:04.500
Find the values of the following at x = 0.
00:29:04.500 --> 00:29:13.400
Here we are going to form the derivatives and then just use these values, plug them in.
00:29:13.400 --> 00:29:18.000
Let us start with ddx(uv).
00:29:18.000 --> 00:29:30.700
Ddx(uv) that is equal to u’ v + u v’.
00:29:30.700 --> 00:29:43.900
U’ is -2, v is -3, this is all at 0.
00:29:43.900 --> 00:29:52.700
+u at 0 which is 7 × v’ which is 2.
00:29:52.700 --> 00:30:03.000
This is 6 + 14 = 20.
00:30:03.000 --> 00:30:04.700
I do not even need to know what the functions are.
00:30:04.700 --> 00:30:08.400
I know what the values of the functions are and their derivative.
00:30:08.400 --> 00:30:13.300
The rest is very simple, good.
00:30:13.300 --> 00:30:22.700
Let us do the ddx(u/v).
00:30:22.700 --> 00:30:39.800
The ddx(u/v) is v u‘ – u v’/ v².
00:30:39.800 --> 00:30:57.200
V is -3, u’ is -2, - u which is 7, × v’ which is 2/ v².
00:30:57.200 --> 00:31:02.400
V is -3².
00:31:02.400 --> 00:31:08.400
You get 6 -14/9.
00:31:08.400 --> 00:31:11.500
6 – 14, I think is -8.
00:31:11.500 --> 00:31:22.300
You get -8/9, good.
00:31:22.300 --> 00:31:29.500
We have ddx of, this time we have v/u.
00:31:29.500 --> 00:31:42.300
This is the denominator × the derivative of the numerator – the numerator × the derivative of the denominator / the denominator².
00:31:42.300 --> 00:31:54.000
Plug the values in, we are going to get 7 × 2 -3 and -2/7².
00:31:54.000 --> 00:32:08.500
That is going to equal 14 – 6/ 49, that equals 8/49.
00:32:08.500 --> 00:32:17.400
We have ddx (5v – 4u).
00:32:17.400 --> 00:32:49.000
That is 5v’, the derivative of this is just the constant × the derivative of this, that is just v’ – 4 × u’ = 5 × 2 – 4 × -2 = 10 + 8 = 18.
00:32:49.000 --> 00:33:02.700
The final one, ddx (u⁴), u is a function of x.
00:33:02.700 --> 00:33:13.900
It is equal to 4 × u³ × du dx which is u’.
00:33:13.900 --> 00:33:23.300
We have 4 × 7³ × -2.
00:33:23.300 --> 00:33:33.000
We end up with -2744.
00:33:33.000 --> 00:33:42.500
We actually have our final one, ddx(7v⁻³).
00:33:42.500 --> 00:33:53.400
V is a function of x, this becomes 7 × -3 × v -4.
00:33:53.400 --> 00:34:02.500
-3 – 1 is -4, × dv dx which is v’.
00:34:02.500 --> 00:34:07.600
V is a function of x, we have to multiply by v’.
00:34:07.600 --> 00:34:09.300
That is the rule for powers.
00:34:09.300 --> 00:34:24.500
This becomes -21/v⁴, I just went ahead and put the -4 down there, × v’.
00:34:24.500 --> 00:34:33.600
That equals -21/-3⁴ × 2.
00:34:33.600 --> 00:34:44.900
You end up getting a final answer of -42/81, if I have done my arithmetic correctly.
00:34:44.900 --> 00:34:50.900
Hope that made sense.
00:34:50.900 --> 00:34:56.800
Let us see, example number 7.
00:34:56.800 --> 00:35:03.600
Find the equation for the tangent to the curve x²/ x³ + 3 at x = 2.
00:35:03.600 --> 00:35:11.900
Very simple, we find the derivative, we put the value 2 into that derivative, that gives us a slope.
00:35:11.900 --> 00:35:18.100
We find the point, the y value, and then we just form the equation.
00:35:18.100 --> 00:35:21.800
Let us go ahead and start with y’.
00:35:21.800 --> 00:35:30.100
Let us see, y’ = this is f and this is g.
00:35:30.100 --> 00:35:35.500
This × the derivative of that – that × the derivative of this/ this².
00:35:35.500 --> 00:35:58.300
We have x³ + 3 × the derivative of this which is 2x – x² × the derivative of this which is 3x²/ x³ + 3².
00:35:58.300 --> 00:36:28.600
Y’ is equal to 2x⁴ + 6x² – 3x⁴/x³ + 3², y’ at 2.
00:36:28.600 --> 00:36:41.400
When I put 2 into this, I end up with -16 + 24 / 121.
00:36:41.400 --> 00:36:46.000
I end up with 8/121.
00:36:46.000 --> 00:36:47.500
This is going to be our slope.
00:36:47.500 --> 00:36:49.500
It is going to be a slope of our line.
00:36:49.500 --> 00:37:00.900
The derivative at the given point is the slope of the tangent line through that point, through the x value of that point.
00:37:00.900 --> 00:37:04.000
The slope is equal to 8/ 121.
00:37:04.000 --> 00:37:08.300
Let us find f(2), we need to find the y value.
00:37:08.300 --> 00:37:22.900
The f(2) that is just going to be 2²/ 2³ + 3 which is going to be 4/11.
00:37:22.900 --> 00:37:33.100
Our point is 2 and 4/11.
00:37:33.100 --> 00:37:45.100
Therefore, our equation is y – 4/11 = 8/121 × x – 2.
00:37:45.100 --> 00:37:53.000
There you go, we want the equation of the tangent to the curve at x = 2.
00:37:53.000 --> 00:37:55.600
When x = 2, the y value is 4/11.
00:37:55.600 --> 00:37:59.000
Another line passes through that point.
00:37:59.000 --> 00:38:02.500
The slope of the line is the derivative at that point.
00:38:02.500 --> 00:38:11.500
Y -y1 = m × x - x1.
00:38:11.500 --> 00:38:13.900
Let us see here.
00:38:13.900 --> 00:38:19.300
X²/x – 7, find d² y dx².
00:38:19.300 --> 00:38:22.100
We need to differentiate this twice.
00:38:22.100 --> 00:38:25.800
Let us start with y’.
00:38:25.800 --> 00:38:46.900
This is f and this is g, this × the derivative of that, which is going to be x -7 × 2x – x² × the derivative of that which is 1/ x – 7².
00:38:46.900 --> 00:39:01.200
This is going to equal 2x² - 14x – x²/ x – 7².
00:39:01.200 --> 00:39:25.100
Therefore, we have a final answer of y’ is equal to 2x² - x², that is just equal to x² – 14x/ x – 7².
00:39:25.100 --> 00:39:29.700
Now we need to find y” which means we differentiate this.
00:39:29.700 --> 00:39:41.700
Let me rewrite this as, y’ is equal to x² - 14x.
00:39:41.700 --> 00:39:46.400
I can do this via the quotient rule again or I can do the product rule.
00:39:46.400 --> 00:39:51.800
I’m just going to go ahead and rewrite this as, x-7⁻².
00:39:51.800 --> 00:40:02.100
I just brought it up, I’m going to do it as a product rule instead, y” which is the derivative of this.
00:40:02.100 --> 00:40:04.100
Let us go to black here.
00:40:04.100 --> 00:40:10.300
This is our f and this is our g.
00:40:10.300 --> 00:40:14.900
Our g is actually this whole thing.
00:40:14.900 --> 00:40:20.900
But notice that g itself is made up of a function raised to a power.
00:40:20.900 --> 00:40:23.200
You just have to be really careful here.
00:40:23.200 --> 00:40:26.900
This is the product rule.
00:40:26.900 --> 00:40:29.200
This is f × g.
00:40:29.200 --> 00:40:36.900
We are going to have f’g + f g’.
00:40:36.900 --> 00:40:44.900
The derivative of this × that + the derivative of this × that.
00:40:44.900 --> 00:41:00.000
F‘ is equal to 2x × g which is x – 7⁻² + f.
00:41:00.000 --> 00:41:09.400
F is equal to x² -14x × the derivative of g’.
00:41:09.400 --> 00:41:24.800
The derivative of this whole thing is -2 × x-7⁻³ × the derivative of what is in here which is just 1.
00:41:24.800 --> 00:41:39.000
Now I simplify, I’m going to write this as 2x/ x – 7² +, this is -, I’m sorry.
00:41:39.000 --> 00:41:45.800
This is going to end up being a -.
00:41:45.800 --> 00:41:50.200
-2 × x² – 14x.
00:41:50.200 --> 00:41:53.300
I’m going to bring this thing down to the denominator.
00:41:53.300 --> 00:41:59.300
I’m going to write it as x – 7³, that makes sense, -2.
00:41:59.300 --> 00:42:02.500
This thing stay on top, I just brought this down.
00:42:02.500 --> 00:42:07.600
I’m going to find a common denominator here by multiplying the top and bottom by x-7
00:42:07.600 --> 00:42:13.600
because I want the denominator to be x -7³.
00:42:13.600 --> 00:42:38.600
This equals 2x × x – 7 – 2 × x² – 14x/ the common denominator which is x – 7³.
00:42:38.600 --> 00:42:59.400
2x² – 14x -2x² + 28x, x-7³.
00:42:59.400 --> 00:43:05.600
2x² – 2x², they go away.
00:43:05.600 --> 00:43:10.100
-14 + 28, I’m left with 14x on top.
00:43:10.100 --> 00:43:16.700
I’m left with x – 7³ in the bottom.
00:43:16.700 --> 00:43:19.800
This is my y”.
00:43:19.800 --> 00:43:23.300
Here is my y’, my first derivative.
00:43:23.300 --> 00:43:27.800
Here is my second derivative.
00:43:27.800 --> 00:43:34.400
I’m going to do this again but this time I’m going to use the quotient rule instead of the power rule,
00:43:34.400 --> 00:43:38.700
just for the sake of seeing another path.
00:43:38.700 --> 00:43:42.700
Let us go ahead and write y’ again.
00:43:42.700 --> 00:43:53.300
Y’ that was equal to x² - 14x/ x -7².
00:43:53.300 --> 00:43:57.900
I’m going to the derivative of this.
00:43:57.900 --> 00:44:03.900
Y” = this × the derivative of that.
00:44:03.900 --> 00:44:18.700
X – 7² × the derivative of this which is 2x – that × the derivative of that x² – 14x × the derivative of this
00:44:18.700 --> 00:44:36.400
which is 2 × x – 7 × the derivative of what is inside which is 1/ x -7².
00:44:36.400 --> 00:44:41.000
That is going to equal, I’m going to multiply it all out.
00:44:41.000 --> 00:44:57.300
X² – 14x + 49 × 2x – 2 ×, I’m going to multiply this all out.
00:44:57.300 --> 00:45:00.200
The x² -14x × x – 7.
00:45:00.200 --> 00:45:02.300
I pull the 2 out, brought it here.
00:45:02.300 --> 00:45:20.000
This is going to be x³ – 7x² -14x² + 98x/ x – 7².
00:45:20.000 --> 00:45:25.700
I’m sorry, x – 7²², this is going to be⁴, my apologies.
00:45:25.700 --> 00:45:28.500
There you go, I’m telling you.
00:45:28.500 --> 00:45:33.700
Vigilance is the most important thing in calculus.
00:45:33.700 --> 00:45:36.800
Let me see, what do we got.
00:45:36.800 --> 00:45:44.800
Let us go ahead and multiply this out.
00:45:44.800 --> 00:46:04.000
We got 2x³ -28x² + 98x – 2x³
00:46:04.000 --> 00:46:22.800
+ 14x² +28x² – 196x/ x – 7⁴.
00:46:22.800 --> 00:46:32.800
2x³ goes with 2x³ – 28x² + 28x².
00:46:32.800 --> 00:46:49.300
I end up with 14x², 98x -, this is going to be -98x/ x – 7⁴.
00:46:49.300 --> 00:46:55.300
I’m going to pull out a 14x and I’m going to write that as x – 7.
00:46:55.300 --> 00:46:58.300
This is x -7⁴.
00:46:58.300 --> 00:47:00.500
This one cancels one of this.
00:47:00.500 --> 00:47:04.800
I’m left with 14x/ x – 7³.
00:47:04.800 --> 00:47:06.500
It is absolutely your choice.
00:47:06.500 --> 00:47:11.100
You can either go product rule, you can go quotient rule.
00:47:11.100 --> 00:47:12.300
You are going to end up with the same answer.
00:47:12.300 --> 00:47:15.800
One of them is more tedious, usually than the other.
00:47:15.800 --> 00:47:21.800
In this particular case, it looks like the quotient rule is the one that is more tedious.
00:47:21.800 --> 00:47:22.900
I hope that helped a little bit.
00:47:22.900 --> 00:47:25.200
Thank you so much for joining us here at www.educator.com.
00:47:25.200 --> 00:47:25.000
We will see you next time, bye.