
#1
Feb1808, 12:14 AM

P: 7

Find y' = dy/dx for x3 + y3 = 4
Okay, now what's really confusing me is that for the y3 is that you need to use the chain rule for it. When you do, the answer is 3y2(dy/dx). How does that actually work? And if anyone can give me any good advice on any good guidelines on how to properly implicitly differentiate, it would be most helpful. Thanks. :) 



#2
Feb1808, 12:19 AM

P: 1,757




#3
Feb1808, 01:14 AM

P: 7

My next question is, in my book, they differentiate d/dx(100xy) and turn it into:
100[x{dy/dx) + y] How did they get to that? 



#4
Feb1808, 01:25 AM

P: 188

Implicit differentiation question
Product rule. You can't just think of y as a constant when you differentiate implicitly.




#5
Feb1808, 01:38 AM

P: 7

Yeah, I felt it had something to do with the product rule, but why did they take out the 100 like that?




#6
Feb1808, 01:47 AM

P: 188

It's just a constant, so that can be moved out. d/dx[100xy] = 100 d/dx[xy] = 100[x*dy/dx + dx/dx*y]




#7
Feb1808, 03:12 AM

P: 7

But can't you do that only if the 100 is being multiplied in both x and y? I mean, the way it is, wouldn't it only be multiplying to either the X or the y, not twice to both of them?




#8
Feb1808, 05:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

You really have to know algebra in order to do calculus! 100xy= (100x)y= x(100y)= 100(xy).




#9
Feb1808, 04:38 PM

P: 7

No, I know that's how it works, but I guess my question is, why isn't the derivative of 100 being taken as well? Why is it being left out?
Sorry for the dumb questions. :/ 



#10
Feb1808, 05:55 PM

P: 1,757

[tex]\frac{d}{dx}(100x)[/tex] What is your answer? Now go back to your derivative properties in which, [tex]\frac{d}{dx}cf(x)=cf'(x)[/tex] Now do you see why it can be left out? 



#11
Feb1808, 06:06 PM

P: 7

Ah, of course! Me so dumb. Thanks a lot. :)



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