## Study23 2 years 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 STEP-BY-STEP PLEASE!!!!!!!!!!!! Thank you so much in advance!!!!!

1. Study23

(I tried solving this problem, but I got a drastically incorrect problem! D: )

2. Study23

3. ivanmlerner

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?

4. Study23

Ummmm...... $$\ \Large 3x^2+2xy+8x=0?$$

5. Study23

@ivanmlerner

6. ivanmlerner

Ok, everything is correct but the second term. Now, remember that y is a function of x and therefore cannot be considered a constant.

7. Study23

Ok... So now what do I do

8. ivanmlerner

Try to diferentiate the second term considering that, do you remember what you do when you are diferentiating a multiplication of functions?

9. Study23

Product Rule? But that's where I get stuck with this problem. HOW do I do that?

10. phi

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 $\frac{d}{dx} x^2y= x^2 \frac{dy}{dx} + 2xy$

11. Study23

Hmmm. So the derivative of $$\ \Large x^2y \text{ is } x^2\frac{dy}{dx}+2xy \text{ ?}$$

12. Study23

Wouldn't that be $$\ \Large 2x+2xy \text{ ?}$$

13. phi

yes, I hope you are able to see how to get it.

14. Study23

I was able to follow along, thank you for that explanation @phi. Now, how do I proceed solving the original problem?

15. phi

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?

16. Study23

I haven't finished the problem, Ill try to solve it now..

17. Study23

I get $$\ \Large \frac{dy}{dx} = -\frac{3}{16x^2y}$$

18. phi

How?!

19. phi

remember $\frac{d}{dx} x^2y= x^2 \frac{dy}{dx} + 2xy$

20. Study23

But the correct answer is: $$\ \Large y'=\frac{-x(3x+2y)}{x^2+8y} .... \text{ I don't know what I did wrong!!! :(}$$

21. Study23

Here's what I did: |dw:1352765472314:dw|

22. Study23

=0.

23. Study23

And then, I:

24. Study23

$3x^2+ 2xy+x^2y'+8x=0,$ $2xy+x^2y'+8x=-3x^2,$$2xy+x^2y'=\frac{-3x}{8}$ $x^2y'=\frac{3x}{8\times2xy}$ $\text{So, } y'=\frac{-3x}{8\times2xy\times x^2}$ $\text{which equals:} \frac{dy}{dx} \frac{-3}{16x2y}$

25. Study23

That 2 in the denominator is an exponent

26. phi

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.

27. Study23

Why $$\ x^2 \frac{dy}{dx} ?$$

28. phi

Your algebra is "suspect". You should be adding or subtracting terms not dividing

29. phi

Why $$x^2dy/dx?$$ Review the post up above where I take the derivative of x^2 y

30. phi

31. Study23

Could I use y' in place of dy/dx? Also, so the product rule is the reason why it is x^2 dy/dx???

32. phi

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

33. Study23

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!

34. phi

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

35. Study23

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