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anonymous
 one year ago
I was given an example of how to derive a derivative from an implicit function, but I don't understand how they got their answer. (will post below)
anonymous
 one year ago
I was given an example of how to derive a derivative from an implicit function, but I don't understand how they got their answer. (will post below)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y^4+3y4x^3=5x+1\] The answer was \[y^4y'+3y'12x^2=5\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The one thing that confuses me is the y's. Are you supposed to add another y variable?

phi
 one year ago
Best ResponseYou've already chosen the best response.2they are taking the derivative with respect to x use the chain rule \[ \frac{d}{dx} \left( y^4+3y4x^3=5x+1 \right) \] \[ \frac{d}{dx} y^4 + 3 \frac{d}{dx} y 4 \frac{d}{dx} x^3 = 5 \frac{d}{dx} x + \frac{d}{dx} 1\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2\[ \frac{d}{dx} y^4= 4 y^3 \frac{d}{dx} y \\ 4y^3 \frac{dy}{dx} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0This is the chain rule. Pretend y (which is a function of x) is f(x). Then whenever you see something like: \[(y^7)' = 7y^6 y'\] This is the same as: \[((f(x))^7)' = 7(f(x))^6*f'(x)\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2another example \[ 4 \frac{d}{dx} x^3= 12 x^2 \frac{dx}{dx} \\ =12x^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well your other example with 'x' makes sense, but are you supposed to just use the power rule for x and us the chain rule for a y variable?

phi
 one year ago
Best ResponseYou've already chosen the best response.2you always use the chain rule. But when you take the derivative with respect to x you get a dx/dx that "goes away" (see example with x^3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there a reason why it goes away for x and not y?

phi
 one year ago
Best ResponseYou've already chosen the best response.2another example: \[ \frac{d}{dx} (x+1)^2 = 2(x+1) \frac{d}{dx} (x+1) \\ = 2(x+1) \left(\frac{d}{dx} x+\frac{d}{dx} 1 \right) \] and that becomes \[ 2(x+1) (1 +0) = 2(x+1) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2I think of it like this: \[ \frac{d}{dx} x = \frac{dx}{dx} = 1\\ \frac{d}{dx} y= \frac{dy}{dx} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Kind of unrelated but amusing, if you treat x as a function and differentiate it with the chain rule, you will get an infinite 'chain': \[x' = 1*x^0 x'=1*x^0*1*x^0*x' =1*x^0*1*x^0*1*x^0*x' \] haha I was hoping this would give some clever way of proving x'=1 but I don't think we can do that here, oh well.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's interesting to think about. Okay here's an example. Tell me if I'm right...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2y^7 \rightarrow 14y^6y'\] and \[2x^3 \rightarrow 6x^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry guys. Openstudy just crashed and reloaded itself randomly. Weird? Anyways, just to make sure, is the "extra y" always added on when taking the derivative of a y variable?

phi
 one year ago
Best ResponseYou've already chosen the best response.2if we take the derivative with respect to a third variable, such as t we would do \[ \frac{d}{dt} (x^2 + y^2) = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] notice the "prime notation" y' leaves out information (we have to remember that it means dy/dx) also, you sometimes see Newton's notation, where we put a dot over the variable to denote the derivative with respect to t (time) \[ \frac{d}{dt} (x^2 + y^2) = 2x \dot{x} + 2y \dot{y} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am sorry, but I am completely lost.

phi
 one year ago
Best ResponseYou've already chosen the best response.2***Is there a reason why it goes away for x and not y?*** you write the ratio of two "infinitesimals" dy/dx if the top and bottom are both dx as in dx/dx that ratio is 1

phi
 one year ago
Best ResponseYou've already chosen the best response.2***I am sorry, but I am completely lost.*** do you know how to use the chain rule?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, but the y and x dilemma was something never explained to me.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh! I think I'm starting to understand now!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so because it's dy/dx it would keep the y variable and like you said dx/dx = 1 so the x would just turn into 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Finally I get it! :D And it would do that with any variable right? dt/dx = keep the t da/dx= keep the a

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I sorry. At times it takes me a minute to catch on. Thank you so much for your help. As well as Empty.
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