WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com and welcome back to AP calculus.
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Today, I thought we would do some more chain rule example problems.
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The chain rule was one of those things that you are pretty much going to be using all the time in your differentiation.
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You are never going to be dealing with straight simple functions.
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It can be a little bit deceiving, in the sense that you think you have a grasp on it.
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And yet somehow things tend to slip away because if you forget to take that last derivative or whatever it is.
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In any case, I just thought it would be nice to do some more example problems.
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Let us jump right on in.
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Nothing particularly difficult, just more practice, something nice to see.
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Differentiate f(x) = sin of cos(tan(x)).
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We have the sin, the cos, the tan.
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We have 3 things, we are going to end up with 3 terms in our derivatives.
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That is how it works.
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The number of things that you actually have, the number of actual functions,
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that is how many terms you end up with in your derivative.
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A good way to think about it if you lose your way.
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I will stick with black or blue, we will stick with blue.
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F’(x) = we are going to take the sin, the derivative of the sin.
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We are working outside in.
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It is going to be the cos of what is inside.
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What is inside is the cos(tan(x)).
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Now we take the derivative of what is inside.
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The derivative of the cos(tan(x)) is –sin tan(x).
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Now we take the derivative of the tan(x) which is sec² x.
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That is it, we are done.
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F’(x), if you want to go and bring this negative sign out front, you are going to end up with f’(x) = -cos of the cos(tan(x)) × sin(tan(x)) × sec² x.
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Is there a way to simplify that, maybe or maybe not?
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That it is not really worth it.
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You are just going to go ahead and use the function as is.
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Nothing particularly strange, you just have to follow through and make sure you go down the line, go down the chain.
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Example number 2, find an equation for the line tangent to the given curve at the given point.
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Our function is sin(x)² π/2.62438.
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Let us go ahead and find the derivative.
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F’(x) is equal to, the derivative of sin is cos.
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It is going to be cos(x)² × the derivative of what is inside, 2x.
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Therefore, if I want I can that 2, it does not really matter.
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Now we are going to go ahead and put π/2 into the x, to actually go ahead and find what this slope is.
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The f’(π/2) = cos(π/2)² × 2 × π/2.
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You end up with, these two cancel here.
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You end up with cos.
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Some number I went ahead and converted it to degrees, just for the heck of it.
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141.7°, it is just going to be π²/ 4.
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The cos(π²)/ 4 is just going to give you some number, and then × π.
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When we solve that, we end up with -0.78 π.
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This is our slope, we want the equation of the line, that is just y - y1 which is 0.6243 = m,
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the slope -0.785 × x - x1, which is –π/2.
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There you go, you can leave it like that, not a problem.
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Example number 3, F(x) is equal to f(g(x)).
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They give a certain values f(4), f(6), f’(4), g(6), g’(6), find f’(6).
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F(x) is f(g(x)), therefore, f’(x) by the chain rule is f’ at g(x) × g’(x).
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Therefore, f’ at 6 = f’ at g(6) × g’(6).
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This equals, g(6) is equal to 4.
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This is going to be f’(4) × g’(6).
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This is f’(4) is equal to 2 and g’(6) is equal to 9.
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Our final answer is 18.
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That is it, just working it out, plugging it in.
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Nice and straightforward.
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Differentiate this function and then graph both the function of the derivative in the same window.
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I just thought that I will throw that extra graphing thing in there.
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Let us see what we have got.
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We have f(x) = this.
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It looks like we are going to be doing the quotient rule.
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This × the derivative of that - that × the derivative of this divided by this².
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F’(x) is going to equal this × the derivative of that.
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That is going to be x × ½ × 1 – x³⁻¹/2 × the derivative of this thing, we still have to take the derivative of that.
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That is going to be × -3x², this × the derivative of that.
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Yes, very good, - that × the derivative of that.
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The derivative of this is 1.
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It just stays 1 – x³¹/2.
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And then, × the derivative of that which is 1.
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We will take all of that /x².
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I’m going to go ahead and rewrite this.
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I will do -3x².
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I’m going to put this and I'm going to go ahead and bring this down to the bottom.
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I’m going to put these, it is going to be -3x³ 3x² x divided by, I will leave the 2.
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I’m going to bring this down below.
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I’m going to actually rewrite it as a radical.
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This is going to be √1 – x³.
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This is going to be –√1 – x³.
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All of that is going to be /x².
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I get myself a common denominator here.
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The common denominator is going to be 2 × 1- √x³.
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We get -3x³ – 2 × 1 – x³.
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Because when I multiply this × that, the radical sign goes away.
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All over 2 × √1 – x³/ x².
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Here we are going to have – 3x³.
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A – and – x³, this is going to be -2x³ – 2/ 2 × √1-x³/ x².
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I will just go ahead and bring this x² and put it in the denominator.
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I’m left with -2x³ – 2.
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If I did my arithmetic correctly, which is always the task up.
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2x², 1 – x³.
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There you go, this is our derivative.
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If I did the algebra correct.
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As far as what it looks like, it basically just looks like this.
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This red one, this is the original f(x).
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This little light blue one, this is f’(x).
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F(x), this is f(x), f’(x) and f’(x), there you go.
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This way, you see the slope is positive.
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If it is positive, it is above the x axis.
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At some point, it actually hits 0, becomes horizontal.
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And then, the slope becomes negative, that is the derivative of the line.
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Here the slope is negative, starts to become a little positive.
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It zeros out again and it actually drops back down.
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It starts to get a little positive but then it drops back down.
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That is it, straight looking graph but these are the functions that we are going to be dealing with in the real world.
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Differentiate f(x) = x – 8/ x + 3⁴.
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Here, f’(x), chain rule.
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We will do 4 × x – 8/ x + 3³ ×, the derivative of what is inside.
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The derivative of what is inside is going to be quotient rule.
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It is going to be x + 3 × the derivative of the top which is 1, -x – 8 × the derivative of the bottom which is 1, all over x + 3².
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This is going to equal, let me bring it down here.
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This is going to equal 4 ×, I‘m going to separate this one out.
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It is going to be 4 × x³/ x + 3³ ×,
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Here we have x + 3 – x – 8.
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I will just write it out, it is not a problem.
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It is going to be x + 3 – x + 8.
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Sorry about that, it is going to be –x.
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A – and -8 is going to be +8.
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This is going to be x + 3, it is going to be².
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This is going to equal, the x – x, they go away.
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We have 3 and we have 8.
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3 + 8 is 11, 11 × 4 is 44.
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On the top we have 44 and we have x – 8³.
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The bottom x + 3³, x + 3².
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We get x + 3⁵, there we go.
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Just have to follow the chain.
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F(x) = sec² 12x.
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This one is really nice and simple.
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Let us see, sec² 12x.
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We have to deal with this one first.
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This is the same as, the sec(12x)².
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We have to do chain rule first.
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It is going to be 2 sec 12x.
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Now we take the derivative of the sec which is going to be sec 12x tan 12x.
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And then, × the derivative of the 12x, there is that.
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Our final answer is going to be, I will take the 2 and the 12 together, it is going to be 24.
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Sec 12x, I will put those together.
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Sec² 12x tan 12x, that is our final answer.
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We have three things going on.
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We have the power function, we have the sec function, and we have the 12x function.
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There are 3 three things so we are going to have 3 terms in our derivative.
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Here is the first term, that takes care of the second term.
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This takes care of the third term.
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Just keep it that way.
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If you want, you can draw the lines in between to separate your terms.
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Identify how many actual functions you have, and then that the terms.
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If you have counted 1, 2, 3, and then you only ended up with two, you know that you are going to need one more.
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That one more is going to be the derivative of that.
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As many functions as you have in your main function, that is how many terms
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you are going to have in your derivative of the function.
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Differentiate this, e ⁺x² + 3x × sin(x²) + 3x/ cos x.
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This is going to be long and painful.
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Maybe not too bad, let us see what is going on here.
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Let us go back to blue.
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We have f’(x), probably going to run out of room but it is not a problem.
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It is going to be this × the derivative of that – that × the derivative of this/ that².
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This × the derivative of this.
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We have cos x × the derivative of this.
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The derivative of this, the top is going to be product rule.
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It is going to be this × the derivative of that + that × the derivative of this.
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Here is what we end up getting.
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It is going to be e ⁺x² + 3x × the derivative of that.
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The derivative of that is going to be cos(x²) + 3x × the derivative of what is inside
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which is going to be 2x + 3 + that × the derivative of this.
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It is going to be + sin(x²) + 3 × the derivative of this.
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The derivative of that is equal to e ⁺x² + 3x ×, I will just bring it down here, the derivative of that which is 2x + 3.
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That is the first part, this × the derivative of that.
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From that, this is still the numerator here, -this × the derivative of this.
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It is going to be –e ⁺x² + 3x × sin(x²) + 3x × the derivative of this × -sin x.
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It is going to be all of this /that².
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What are we going to do with this?
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Probably, you do not want to simplify it too much.
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I’m going to go ahead and pull out a, 2x + 3, I have a 2x + 3, I’m going to pull that out.
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I have an e ⁺x² + 3x, that is one term.
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I have an e ⁺x² + 3x, that is another term.
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I have an e ⁺x² + 3x, that is another term.
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Let us see, let me just take this first part here.
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This first part, let me deal with that.
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I’m going to go ahead and factor out a 2x + 3.
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I’m going to factor out an e ⁺x² + 3x.
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I’m also going to factor out a cos(x).
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That is going to be cos(x)² + 3x + sin(x)² + 3x.
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I’m going to deal with this term right here.
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I’m just going to actually leave it alone, - and – becomes +.
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It becomes + e ⁺x² + 3x × sin(x)² + 3x × sin(x).
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All of that is the numerator, and then, cos² x.
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I think it is best to just leave it at that point, just one little step of simplification.
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You could have left it at that, it is not a big deal.
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Just make it a little bit cleaner, if nothing else.
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There you go, very complicated as we would expect.
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We have a very complicated looking function.
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Exponential, trigonometric, there is going to be a lot going on.
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Differentiate f(x) = 6 ⁺sin(π/2x).
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Let us recall, anytime we have a constant raised to some power,
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the derivative of that = a ⁺u and then the nat-log of the base which is a.
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F’(x) that is going to equal 6 ⁺sin(π/2x) × ln (6) × du dx, chain rule.
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Now we take the derivative of this.
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The derivative of sin(π/2x) is cos(π/2x) × the derivative of π/2x which is π/2.
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We can put some things together here.
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Let me go back to blue.
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Ln (6) is a constant, π/2 is a constant.
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Let us write it as π ln 6/ 2 × this thing × this thing ×,
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Let me go ahead and put a parenthesis.
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6 ⁺sin(π/2x) × cos(π/2x).
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There we go, that is our final answer.
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This one right here, the rate of change of the radius of a spherical balloon with respect to time as we inflate it is 1.5 cm/s,
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find an expression for the rate of change of the volume of the balloon with respect to time.
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The rate of change of the radius of the spherical balloon with respect to time, that is dr dt, that is equal to 1.5 cm/s.
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Basically, you have this balloon, you are inflating it, it is getting bigger and bigger.
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This radius is changing, for every second it is growing by 1.5 cm.
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That is all this says.
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It is the cm/s, radius is in centimeters, time is in second.
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This is a rate of change, that is how we express the rate of change.
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Dy dx is the rate of change of y for every unit change in x.
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For every unit interval of time, 1 second, the radius changes by 1.5 cm.
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That is all this is, it is just the rate.
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They want to know, find an expression for the rate of change of the volume with respect to time.
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What they want is dv dt.
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We have a relationship between v and r.
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Let us go ahead and deal with that.
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We know that the volume of a sphere is 4/3 π r³.
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We have dr dt, we want dv dt.
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Dv dt, we will just differentiate everything with respect to t.
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Dv dt = 4/3 π ×, r is a function of t.
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We treat it chain rule, it is going to be × 3 r² dr dt.
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The 3 and the 3 cancel, therefore, I get dv dt = 4 π r² dr dt.
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Dr dt is 1.5, I plug that in here.
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I get 4 π r² × 1.5.
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The final answer is, the rate of change of volume with respect to time is equal to 6 π r².
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Whatever the radius happens to be at moment, I put that into this equation.
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That will tell me, at that moment, how fast the volume is changing.
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Notice, here, the rate of change of the radius is constant.
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That means the radius is growing by 1.5 cm every second, it is not changing.
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But every second it is 1.5 cm, another second 1.5 cm, another second 1.5 cm.
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The volume is not linear, it is not constant.
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It actually depends on what r is.
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As r increases, the rate at which the volume is changing actually keeps getting more and more.
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Because now, the rate of change of volume depends on r.
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As r changes, the square of r, the number is going to change.
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If r is 4, this is going to be 4 × 4 is 16, 6 × 16 π.
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When the radius is 4, the volume is changing at 6 × 16 π.
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If the radius is 10, 10² is 100, 6 × 100.
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When radius is 100, the volume is changing per second at 600 π.
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The rate of change of the radius is constant but the rate of change of volume is not constant.
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It actually depends on r.
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This is the general procedure.
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You will be seeing a lot of this when we do related rates.
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I think that is actually the last example for today.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.