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Okay, so here's the thing. I'm REALLY REALLY CONFUSED BY IMPLICIT DIFFERENTIATION!!! The FUNCTION: \(\ \large x^3+x^2y+4x^2=6 \). Find \(\ \frac{dy}{dx} \). I've been on this problem for some 45 minutes, and have gotten no where on this homework assignment (this is the first problem). So, PLEASE PLEASE PLEASE HELP ME!!! SHOW me STEPBYSTEP PLEASE!!!!!!!!!!!! Thank you so much in advance!!!!!
 one year ago
 one year ago
Okay, so here's the thing. I'm REALLY REALLY CONFUSED BY IMPLICIT DIFFERENTIATION!!! The FUNCTION: \(\ \large x^3+x^2y+4x^2=6 \). Find \(\ \frac{dy}{dx} \). I've been on this problem for some 45 minutes, and have gotten no where on this homework assignment (this is the first problem). So, PLEASE PLEASE PLEASE HELP ME!!! SHOW me STEPBYSTEP PLEASE!!!!!!!!!!!! Thank you so much in advance!!!!!
 one year ago
 one year ago

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Study23Best ResponseYou've already chosen the best response.1
(I tried solving this problem, but I got a drastically incorrect problem! D: )
 one year ago

ivanmlernerBest ResponseYou've already chosen the best response.0
What implicit diferentiation does, is the same thing as the normal diferentiation. You diferentiate on both sides in respect to one variable, and the derivative of the other variable that depends on the first appears isolated in normal diferentiation, and not isolated in implicit, you only need to isolate it. In your function, you first diferentiate on both sides in respect to x, you can of course isolate y, wich is easy in this function, but lets do it in the way the problem wants to. Diferentiating on both sides, what have you got?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Ummmm...... \(\ \Large 3x^2+2xy+8x=0? \)
 one year ago

ivanmlernerBest ResponseYou've already chosen the best response.0
Ok, everything is correct but the second term. Now, remember that y is a function of x and therefore cannot be considered a constant.
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Ok... So now what do I do
 one year ago

ivanmlernerBest ResponseYou've already chosen the best response.0
Try to diferentiate the second term considering that, do you remember what you do when you are diferentiating a multiplication of functions?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Product Rule? But that's where I get stuck with this problem. HOW do I do that?
 one year ago

phiBest ResponseYou've already chosen the best response.1
Let's do just the \(x^2 y\) as an example. First, we remember the product rule d(u v) = u dv + v du second, we take the derivative with respect to x \[ \frac{d}{dx} x^2y= x^2 \frac{d}{dx}y + y \frac{d}{dx}x^2\] look carefully at this expression \[ \frac{d}{dx}y \text{ is just } \frac{dy}{dx}\] and \[ \frac{d}{dx}x^2 \text { is } 2 x \frac{d}{dx}x= 2x \frac{dx}{dx} = 2x\] we find \[ \frac{d}{dx} x^2y= x^2 \frac{dy}{dx} + 2xy \]
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Hmmm. So the derivative of \(\ \Large x^2y \text{ is } x^2\frac{dy}{dx}+2xy \text{ ?} \)
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Wouldn't that be \(\ \Large 2x+2xy \text{ ?} \)
 one year ago

phiBest ResponseYou've already chosen the best response.1
yes, I hope you are able to see how to get it.
 one year ago

Study23Best ResponseYou've already chosen the best response.1
I was able to follow along, thank you for that explanation @phi. Now, how do I proceed solving the original problem?
 one year ago

phiBest ResponseYou've already chosen the best response.1
Finish the derivative (they are all just x terms so they are what you are used to...) then "solve" for dy/dx First what do you get for the derivate?
 one year ago

Study23Best ResponseYou've already chosen the best response.1
I haven't finished the problem, Ill try to solve it now..
 one year ago

Study23Best ResponseYou've already chosen the best response.1
I get \(\ \Large \frac{dy}{dx} = \frac{3}{16x^2y}\)
 one year ago

phiBest ResponseYou've already chosen the best response.1
remember \[ \frac{d}{dx} x^2y= x^2 \frac{dy}{dx} + 2xy \]
 one year ago

Study23Best ResponseYou've already chosen the best response.1
But the correct answer is: \(\ \Large y'=\frac{x(3x+2y)}{x^2+8y} .... \text{ I don't know what I did wrong!!! :(}\)
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Here's what I did: dw:1352765472314:dw
 one year ago

Study23Best ResponseYou've already chosen the best response.1
\[3x^2+ 2xy+x^2y'+8x=0,\] \[2xy+x^2y'+8x=3x^2,\]\[2xy+x^2y'=\frac{3x}{8}\] \[x^2y'=\frac{3x}{8\times2xy}\] \[\text{So, } y'=\frac{3x}{8\times2xy\times x^2}\] \[\text{which equals:} \frac{dy}{dx} \frac{3}{16x2y}\]
 one year ago

Study23Best ResponseYou've already chosen the best response.1
That 2 in the denominator is an exponent
 one year ago

phiBest ResponseYou've already chosen the best response.1
yes that looks ok \[ 3x^2 +2xy + x^2 \frac{dy}{dx} +8x=0\] you can factor out an x. \[ 3x +2y + x \frac{dy}{dx} +8=0\] move the x dy/dx to the other side \[ 3x +2y + 8= x \frac{dy}{dx} \] divide by x \[ \frac{3x2y8}{x} = \frac{dy}{dx} \] compare to wolfram http://www.wolframalpha.com/input/?i=implicit+diff++x%5E3%2Bx%5E2y%2B4x%5E2%3D6.
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Why \(\ x^2 \frac{dy}{dx} ? \)
 one year ago

phiBest ResponseYou've already chosen the best response.1
Your algebra is "suspect". You should be adding or subtracting terms not dividing
 one year ago

phiBest ResponseYou've already chosen the best response.1
Why \( x^2dy/dx?\) Review the post up above where I take the derivative of x^2 y
 one year ago

phiBest ResponseYou've already chosen the best response.1
Your "correct answer" does not match this problem...
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Could I use y' in place of dy/dx? Also, so the product rule is the reason why it is x^2 dy/dx???
 one year ago

phiBest ResponseYou've already chosen the best response.1
yes y' for dy/dx is ok. just so long as you remember what it means... and yes the product rule is d (u * v) = u * dv + v*du that u*dv means you leave u "alone" and multiply times the derivative of v if we match u with x^2 and v with y, that means you have x^2 * d y
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Okay, I think I get this now... I'll try this problem again. Thank you so much for spending the time with me @phi!!! A medal can not express the utmost appreciation I have for your help!
 one year ago

phiBest ResponseYou've already chosen the best response.1
I notice part of your trouble comes from not doing the algebra correctly. You may need to brush up on it...but post your questions and someone will help
 one year ago

Study23Best ResponseYou've already chosen the best response.1
Okay! You're right! On the past two tests I have lost points for distributive property errors not so much as for the calculus  it's more the algebra
 one year ago
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