anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[y^4+3y-4x^3=5x+1\] The answer was \[y^4y'+3y'-12x^2=5\]
anonymous
  • anonymous
The one thing that confuses me is the y's. Are you supposed to add another y variable?
phi
  • phi
they are taking the derivative with respect to x use the chain rule \[ \frac{d}{dx} \left( y^4+3y-4x^3=5x+1 \right) \] \[ \frac{d}{dx} y^4 + 3 \frac{d}{dx} y -4 \frac{d}{dx} x^3 = 5 \frac{d}{dx} x + \frac{d}{dx} 1\]

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phi
  • phi
\[ \frac{d}{dx} y^4= 4 y^3 \frac{d}{dx} y \\ 4y^3 \frac{dy}{dx} \]
Empty
  • Empty
This 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
  • phi
another example \[ -4 \frac{d}{dx} x^3= -12 x^2 \frac{dx}{dx} \\ =-12x^2 \]
anonymous
  • anonymous
Well 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
  • phi
you 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
  • anonymous
Is there a reason why it goes away for x and not y?
phi
  • phi
another 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
  • phi
I think of it like this: \[ \frac{d}{dx} x = \frac{dx}{dx} = 1\\ \frac{d}{dx} y= \frac{dy}{dx} \]
Empty
  • Empty
Kind 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
  • anonymous
That's interesting to think about. Okay here's an example. Tell me if I'm right...
anonymous
  • anonymous
\[2y^7 \rightarrow 14y^6y'\] and \[2x^3 \rightarrow 6x^2\]
phi
  • phi
yes
anonymous
  • anonymous
.
anonymous
  • anonymous
Sorry 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
  • phi
if 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
  • anonymous
I am sorry, but I am completely lost.
phi
  • phi
***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
  • phi
***I am sorry, but I am completely lost.*** do you know how to use the chain rule?
anonymous
  • anonymous
Yes, but the y and x dilemma was something never explained to me.
anonymous
  • anonymous
Oh! I think I'm starting to understand now!
anonymous
  • anonymous
so 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.
phi
  • phi
yes
anonymous
  • anonymous
Finally I get it! :D And it would do that with any variable right? dt/dx = keep the t da/dx= keep the a
phi
  • phi
yes
anonymous
  • anonymous
I sorry. At times it takes me a minute to catch on. Thank you so much for your help. As well as Empty.
phi
  • phi
yw

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