AP®︎/College Calculus AB
- Implicit differentiation
- Worked example: Implicit differentiation
- Worked example: Evaluating derivative with implicit differentiation
- Implicit differentiation
- Showing explicit and implicit differentiation give same result
- Implicit differentiation review
Worked example: Evaluating derivative with implicit differentiation
Sal finds the slope of the tangent line to the curve x²+(y-x)³=28 at x=1 using implicit differentiation. Created by Sal Khan.
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- at2:25you decide to distribute, Why? could you just put all non dy/dx terms on the opposite side of the equation and then add 1 to both sides making the final equation dy/dx= (1-2x)/3(y-x)^2(47 votes)
- Yes, and that would be considerably easier than what Sal did, I think, but we need to be careful on the algebra. The final equation would actually be (-2x / (3(y-x)^2)) + 1. Looking at your equation, I think you may have added 1 to both sides a bit too soon.(73 votes)
- I'm assuming that simplifying fractions for problems like these is a bad idea. My initial instinct was to cancel out the 27s in the numerator and denominator.(13 votes)
- Simplifying is VERY important in problems like these, but you cannot cancel terms across the fraction line. Before you can cancel something in the numerator and denominator, you need to factor out common factors or else factor out like binomials, which are just complicated factors. Once you have separated out the common factors, then it is fine to cancel.
For example, let's say instead of what Sal got at the end of the video, you have 27x - 9 in the numerator, and 27x in the denominator. You cannot cancel out the 27x because they are both just terms, but you CAN and should factor this hypothetical numerator and denominator as follows:
27x - 9 can be factored by removing the common factor of of 9 leaving. . .
In my denominator, 27x contains 3 ∙ 3 ∙ 3 ∙ x
So it also contains a 9 times 3x
So in the numerator, you have a factor of 9 and in the denominator, there is also a factor of 9
Now you can cancel those factors leaving 3x - 1 in the numerator and 3x in the denominator.(31 votes)
- How do you know when you have to use implicit differentiation? Is it when you are trying to find the rate of change with respect to another function? So if you wanted to know the rate at which f(y) is changing with respect to x with something like x^2y^2=1-- this is when we use this rule?(7 votes)
- if there is more then one variable in the expression you are trying to differentiate, you would use implicit differentiation.(22 votes)
- At about4:20, why would you not simplify by crossing out the 3(y-x)^2's and having -2x?(2 votes)
- You don't want to cancel when sums/differences, only products.(14 votes)
- At the end of the video, I could see why he didn't further simplify expression of dy/dx so that he could show us we do not have to substitute y for an explicit expression of the original equation, he wanted to show us, if we know y and we know x, we can just substitute those into the expression of dy/dx. But, he could 've simplified it to 1-(2x/3(y-x)^2 and still made his point. Wouldn't that be much simpler?(4 votes)
- Yes, that works. It would probably make it a little bit easier to plug in the numbers at the end of the video.(4 votes)
- Okay, so instead of factoring (3(y-x)^2)((dy/dx)-1) instead I moved subtracted 2x from both sides. then I divided 3(y-x)^2 from both sides.
That gave me dy/dx -1 = -2x/(3(y-x)^2). Then I added 1 to both sides.
My final equation looked like: dy/dx = -2x/(3(x-y)^2) -1.
And as hard as I try, I don't understand why that is wrong. As a matter of fact, it seems much easier than factoring 3(x-y)^2. But, when it came to plugging in the coordinate (1,4) I got myself into a muddle because,
dy/dx = -2(1) / (3(4-1)^2) - 1
dy/dx = -2/(3(3)^2) -1
dy/dx = -2/(3*9) -1
dy/dx = -2/27 -1
Obviously that isn't the answer Sal got. So, I was wondering what I am missing and why factoring earlier on changes things so dramatically.(2 votes)
- The sign of -1, since you added it to both sides, it should be +1.(3 votes)
- I suppose the question is redundant since he did it, but can you just multiply the "3(y-x)^2" in its entirity with a term inside the brackets? I've never seen something like that.(3 votes)
- Yes you could, it would just make the equation a lot more complicated. Sometimes it's better to not simplify, because (for me at least) 3(y-x)^2 is a lot easier to read than 3y^2 - 6yx + 3x^2.(1 vote)
- how would you know when to take (dy/dx) of a term and when to take (dx/dx)?(2 votes)
- Taking dx/dx of a function is not a thing as that simply equals 1. You always take the derivative with respect to x of both sides in an implicit relation. Then you use the chain rule to simplify. After that, you bring all the dy/dx terms to one side and the other terms to the other side and then simply solve for dy/dx. Hope this helps!(3 votes)
- When you do d/dx(xy) how does it come out to be y+x*(dy/dx)? I'm confused on what steps you take and how you got the outcome(2 votes)
- The product rule for differentiation has been used.
First we differentiate x with respect to x(which is 1)multiplied by y remaining as it is which turns out to be y*1=y.
This is added to the differentiation of y with respect to x which is clearly the derivative of y dy/dx, which is multiplied by x remaining as it is. That's how we get y+x*(dy/dx). I hope this helps.(3 votes)
- How would you differentiate xy? I'm getting confused trying to work it out. What is the inner function and the outer function?(2 votes)
- First, we note that two functions are multiplied together, meaning that we need to use the product rule. The two functions are the function of x (just x in this case) and the function of y (which is also just y), correct?
So, the product rule says that the derivative of f*g, where f and g are both functions, is f'*g + f*g'. If we apply this to xy and differentiate with respect to x, we have [(d/dx)x * y] + [x * (d/dx)y].
The derivative of x is just 1. The derivative of y with respect to x is slightly more complex. Since y is a function of x, the derivative of y with respect to x is dy/dx, or y' (whichever notation you prefer). If we substitute this in, the final result is: y + xy'.
Hopefully this made sense. If not, feel free to ask any clarifying questions you have. Best of luck as you continue your studies!(2 votes)
We've been doing a lot of examples where we just take implicit derivatives, but we haven't been calculating the actual slope of the tangent line at a given point. And that's what I want to do in this video. So what I want to do is figure out the slope at x is equal to 1. So when x is equal to 1. And as you can imagine, once we implicitly take the derivative of this, we're going to have that as a function of x and y. So it'll be useful to know what y value we get to when our x is equal to 1. So let's figure that out right now. So when x is equal to 1, our relationship right over here becomes 1 squared, which is just 1 plus y minus 1 to the third power is equal to 28. Subtract 1 from both sides. You get y minus 1 to the third power is equal to 27. It looks like the numbers work out quite neatly for us. Take the cube root of both sides. You get y minus 1 is equal to 3. Add 1 to both sides. You get y is equal to 4. So we really want to figure out the slope at the point 1 comma 1 comma 4, which is right over here. When x is 1, y is 4. So we want to figure out the slope of the tangent line right over there. So let's start doing some implicit differentiation. So we're going to take the derivative of both sides of this relationship, or this equation, depending on how you want to view it. And so let's skip down here past the orange. So the derivative with respect to x of x squared is going to be 2x. And then the derivative with respect to x of something to the third power is going to be 3 times that something squared times the derivative of that something with respect to x. And so what's the derivative of this with respect to x? Well the derivative of y with respect to x is just dy dx. And then the derivative of x with respect to x is just 1. So we have minus 1. And on the right-hand side we just get 0. Derivative of a constant is just equal to 0. And now we need to solve for dy dx. So we get 2x. And so if we distribute this business times the dy dx and times the negative 1, when we multiply it times dy dx, we get-- and actually I'm going to write it over here-- so we get plus 3 times y minus x squared times dy dx. And then when we multiply it times the negative 1, we get negative 3 times y minus y minus x squared. And then of course, all of that is going to be equal to 0. Now all we have to do is take this and put it on the right-hand side. So we'll subtract it from both sides of this equation. So on the left-hand side-- and actually all the stuff that's not a dy dx I'm going to write in green-- so on the left-hand side we're just left with 3 times y minus x squared times dy dx, the derivative of y with respect to x is equal to-- I'm just going to subtract this from both sides-- is equal to negative 2x plus this. So I could write it as 3 times y minus x squared minus 2x. So we're adding this to both sides and we're subtracting this from both sides. Minus 2x. And then to solve for dy dx, we've done this multiple times already. To solve for the derivative of y with respect to x. The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x. All of that over this stuff, 3 times y minus x squared. And we can leave it just like that for now. So what is the derivative of y with respect to x? What is the slope of the tangent line when x is 1 and y is equal to 4? Well we just have to substitute x is equal to 1 and y equals 4 into this expression. So it's going to be equal to 3 times 4 minus 1 squared minus 2 times 1. All of that over 3 times 4 minus 1 squared, which is equal to 4 minus 1 is 3. You square it. You get 9. 9 times 3 is 27. You get 27 minus 2 in the numerator, which is going to be equal to 25. And in the denominator, you get 3 times 9, which is 27. So the slope is 25/27. So it's almost 1, but not quite. And that's actually what it looks like on this graph. And actually just to make sure you know where I got this graph. This was from Wolfram Alpha. I should have told you that from the beginning. Anyway, hopefully you enjoyed that.