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Study23
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 STEP-BY-STEP PLEASE!!!!!!!!!!!! Thank you so much in advance!!!!!
(I tried solving this problem, but I got a drastically incorrect problem! D: )
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?
Ummmm...... \(\ \Large 3x^2+2xy+8x=0? \)
Ok, everything is correct but the second term. Now, remember that y is a function of x and therefore cannot be considered a constant.
Ok... So now what do I do
Try to diferentiate the second term considering that, do you remember what you do when you are diferentiating a multiplication of functions?
Product Rule? But that's where I get stuck with this problem. HOW do I do that?
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 \[ \frac{d}{dx} x^2y= x^2 \frac{dy}{dx} + 2xy \]
Hmmm. So the derivative of \(\ \Large x^2y \text{ is } x^2\frac{dy}{dx}+2xy \text{ ?} \)
Wouldn't that be \(\ \Large 2x+2xy \text{ ?} \)
yes, I hope you are able to see how to get it.
I was able to follow along, thank you for that explanation @phi. Now, how do I proceed solving the original problem?
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?
I haven't finished the problem, Ill try to solve it now..
I get \(\ \Large \frac{dy}{dx} = -\frac{3}{16x^2y}\)
remember \[ \frac{d}{dx} x^2y= x^2 \frac{dy}{dx} + 2xy \]
But the correct answer is: \(\ \Large y'=\frac{-x(3x+2y)}{x^2+8y} .... \text{ I don't know what I did wrong!!! :(}\)
Here's what I did: |dw:1352765472314:dw|
\[3x^2+ 2xy+x^2y'+8x=0,\] \[2xy+x^2y'+8x=-3x^2,\]\[2xy+x^2y'=\frac{-3x}{8}\] \[x^2y'=\frac{3x}{8\times2xy}\] \[\text{So, } y'=\frac{-3x}{8\times2xy\times x^2}\] \[\text{which equals:} \frac{dy}{dx} \frac{-3}{16x2y}\]
That 2 in the denominator is an exponent
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 \[ \frac{-3x-2y-8}{x} = \frac{dy}{dx} \] compare to wolfram http://www.wolframalpha.com/input/?i=implicit+diff++x%5E3%2Bx%5E2y%2B4x%5E2%3D6.
Why \(\ x^2 \frac{dy}{dx} ? \)
Your algebra is "suspect". You should be adding or subtracting terms not dividing
Why \( x^2dy/dx?\) Review the post up above where I take the derivative of x^2 y
Your "correct answer" does not match this problem...
Could I use y' in place of dy/dx? Also, so the product rule is the reason why it is x^2 dy/dx???
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
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!
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
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