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. :)
http://en.wikipedia.org/wiki/Implicit_differentiation#Implicit_differentiation [tex]3x^2+3y^2\frac{dy}{dx}=0[/tex]
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?
Yeah, I felt it had something to do with the product rule, but why did they take out the 100 like that?
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?
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. :/
Ok, take the derivative of this problem. Treat it as if it's a product. [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?