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# Worked example: Evaluating derivative with implicit differentiation

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.D (LO)
,
FUN‑3.D.1 (EK)
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.
• 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?
• if there is more then one variable in the expression you are trying to differentiate, you would use implicit differentiation.
• 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.
• You don't want to cancel sum/difference bits. Only cancel when you have the same products in both the numerator and denominator.
• At about , why would you not simplify by crossing out the 3(y-x)^2's and having -2x?
• You don't want to cancel when sums/differences, only products.
• 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?
• Yes, that works. It would probably make it a little bit easier to plug in the numbers at the end of the video.
• 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.
• The sign of -1, since you added it to both sides, it should be +1.
• 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.
• 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)
• 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?
• 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!
• At why didnt he just plug in the x and y which we already knew and then the algebra would become a lot easier?
• I agree Raphael, and when I solved the problem myself before finishing the video, this is exactly what I did. I feel like it was easier than Sal's solution too, since my next step was just:

2 + 27(dy/dx) - 27 = 0

then:
27(dy/dx) = 25

and finally:
dy/dx = 25/27