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Implicit differentiation (advanced example)

Implicit differentiation of y = cos(5x - 3y). Created by Sal Khan.

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  • leaf green style avatar for user dobrien
    In this video at as well as in the last video, sal subtracts something from both sides that contains dy/dx, in order to get all dy/dx to the same side. But I don't get how one of the dy/dx becomes 1. In this video it results in (1-3sin(5x-3y))dy/dx. How does dy/dx-3sin(5x-3y)*dy/dx= (1-3sin(5x-3y))*dy/dx
    (21 votes)
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    • leaf green style avatar for user Lucas Van Meter
      He factored the dy/dx out. For example, in the expression x+5x we can factor out the x to get (1+5)x. The 1 is left behind to show that we had an x there before we factored. Another way to see this is to try multiplying the dy/dx back into the expression. If there was no 1 sitting there than we would have magically lost the dy/dx term in the original expression.
      (42 votes)
  • mr pants teal style avatar for user Diana Dai
    Is this still counted as implicit derivatives? I'm confused because I thought explicit derivatives was where y was isolated on one side. What's the difference between the two?
    (11 votes)
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    • leaf red style avatar for user Yosh Wakeham
      The difference is that we have y terms on both sides of the equation (as y is part of the argument of the cos function). Although we have y on its own on the left-hand side, this is not the equation for y as a function of x. Thus, implicit differentiation is called for.
      (12 votes)
  • blobby green style avatar for user Sandra Reynolds
    Can I treat dy/dx as a fraction?

    Can I multiply both sides by dx/dy to cleat the dy/dx?

    I am not sure why one would do this, I just want to know if dy/dx is now its own thing or is it still a fraction.
    (3 votes)
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  • blobby green style avatar for user Arne Magnus Lorentzen Ulland
    Hello
    I multiply the two paranteses before I multiply with -sin.
    -sin (5x-3y)(5-3dy/dx) = -sin25x + sin 15x*dy/dx + sin15y - 9ydy/dx
    My answer would then be :
    dy/dx = (-sin25x+sin15y)/(-sin15x+1+9y)
    Is this correct? How come Sal uses -5 sin (5x) and not -sin25x.?
    (3 votes)
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    • piceratops ultimate style avatar for user Just Keith
      No. The expression sin(5x-3y) means that you are taking the sine function of (5x-3y), That is not a multiplication and you certainly don't use the distributive property.
      Thus, sin (kx) is NOT sine times kx. sin (kx) = 1/2 {ie^(-kxi)- ie^(kxi)}

      Instead, you treat sin(5x-3y) as a single entity. You can break that up using a trigonometric identity, but that is not a more simple form.
      For reference sake, though:
      sin(5x-3y)=cos(3y)sin(5x)-cos(5x)sin(3y)

      Finally, remember that k( sin x) is NOT equal to sin kx.
      (10 votes)
  • mr pants teal style avatar for user berylliumnile
    At he substracted 3sin(5x-3y) from 1 and at the end he got -5sin(5x-3y) / (1-3sin(5x-3y)). However if we substract the other way and get 3sin(5x-3y)-1 we can get 5sin(5x-3y) / (3sin(5x-3y)-1) . Would it not be right as well?
    (4 votes)
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  • ohnoes default style avatar for user Cyan Wind
    At first, I got confused. However, I came to this conclusion: d/dx is the differential operator (similar to +, -, *, /). dy/dx = d/dx (y) is the variable (similar to a = 5, b = 100, c = -0.12); so we can apply operators to it, right?
    (5 votes)
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  • mr pants teal style avatar for user Shashank Biradar
    I was just curious as to why we must again solve for dy/dx, since that is already what we have once we use the chain rule
    (1 vote)
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    • male robot hal style avatar for user Ben
      The problem is that you had dy/dx on both sides of the equation, and the goal was to find the derivative of y with respect to x. You need the dy/dx isolated for the same reason you don't leave a linear equation as y=2x-y. It makes it much simpler to do any follow up work if you needed the equation if it's already prepared for you. This way when you plug in values for your variables you don't need to do any more algebra, it's just simple order of operations. I hope that clarified everything a little for you.
      (8 votes)
  • spunky sam blue style avatar for user Chunmun
    How to know that the equation given to us will be solved by applying product rule or chain rule ?
    Please explain with the examples if possible .
    Thank you .
    (2 votes)
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    • mr pink red style avatar for user Wei Xia
      most of the time you can use the product rule when you see two variables multiplying eachother. For example like 3x4x^2. Usually a parentnesis is involved when you have to use the Chain Rule. Like (x+6)^9. Expanding it would take too much work, so you use the Chain Rule.
      (6 votes)
  • blobby green style avatar for user harry song
    would the answer -5sin(5x-3y)/4 be right as well?
    (2 votes)
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    • leaf green style avatar for user T Fletcher
      I don't think so. If we plug in a point that we know works for the original equation, say (pi/10, 0) we end up with two different answers for dy/dx: -5/4 for your expression and 5/2 for the expression in the video. However, I don't know if this is a kosher way to test.
      (3 votes)
  • mr pants teal style avatar for user Wiebke Janßen
    At around : Why do you subtract 3 sin(5x-3y) from both sides?
    (2 votes)
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Video transcript

Let's say we have the relationship y is equal to cosine of 5x minus 3y. And what I want to find is the rate at which y is changing with respect to x. And we'll assume that y is a function of x. So let's do what we've always been doing. Let's apply the derivative operator to both sides of this equation. On the left-hand side, right over here, we get dy/dx is equal to-- now here on the right hand side, we're going to apply the chain rule. The derivative of the cosine of something, with respect to that something, is going to be equal to negative sine of that something. So negative sine of 5x minus 3y. And then we have to multiply that by the derivative of that something with respect to x. So what's the derivative of the something with respect to x? Well the derivative of 5x with respect to x is just equal to 5. And the derivative of negative 3y with respect to x is just negative 3 times dy/dx. Negative 3 times the derivative of y with respect to x. And now we just need to solve for dy/dx. And as you can see, with some of these implicit differentiation problems, this is the hard part. And actually, let me make that dy/dx the same color. So that we can keep track of it easier. So this is going to be dy/dx. And then I can close the parentheses. So how can we do it? It's just going to be a little bit of algebra to work through. Well, we can distribute the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going to do that in the yellow color-- we get dy/dx is equal to-- you distribute the negative sine of 5x minus 3y. You get-- so let me make sure we know what we're doing. It's going to be, we're going to distribute that, and we're going to distribute that. So you're going to have 5 times all of this. So you're going to have negative this 5 times the sine of 5x minus 3y. And then you're going to have the negative times a negative, those are going to, you're going to end up with a positive. And so you're going to end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this is essentially a 1 dy/dx. So if we subtract this from both sides, we are left with-- So on the left-hand side, we're going to have a 1 dy/dx, and we're going to subtract from that 3 sine of 5x minus 3y dy/dx's. So you're going to have 1 minus 3-- I'll keep the color for the 3 for fun-- 3 sine of 5x minus 3y dy/dx's on the left-hand side, is going to be equal to, well, we subtracted this from both sides. So on the right-hand side, this is going to go away. So we're just going to be left with a negative 5 sine of 5x minus 3y. And we're in the home stretch now. To solve for dy/dx, we just have to divide both sides of the equation by this. And we are left with dy/dx is equal to this thing, negative 5 times the sine of 5x minus 3y. All of that over 1 minus 3 sine of 5x minus 3y. And we are done.