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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 3

Lesson 3: Implicit differentiation

# 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.

## Want to join the conversation?

• at you 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 •   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.
• 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. •  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. . .
9(3x -1)
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.
• 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? • At about , why would you not simplify by crossing out the 3(y-x)^2's and having -2x? • 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? • 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. • 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. • how would you know when to take (dy/dx) of a term and when to take (dx/dx)? • 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 • 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.
• How would you differentiate xy? I'm getting confused trying to work it out. What is the inner function and the outer function? 