Find the derivative, using implicit differentiation: $$2x^3=(3xy+1)^2$$


Used the chain rule for the derivative $(3xy+1)^2$. Do I move the $2x^3$ over once I get its derivative, which is $6x^2$?

I know the answer is $(-3y^2x-y+x^2)/(x(3yx+1))$, but I am have trouble solving this problem.

  • $\begingroup$ Where exactly is your problem? Differentiate both sides, then solve for $dy/dx$. $\endgroup$ – preferred_anon May 31 '14 at 23:37
  • $\begingroup$ Welcome to math.SE! This question is missing context and details: Please improve it by including your thoughts and efforts on this problem, explaining what you're having trouble with. For example, what derivative rules do you know, and how do they relate this problem? $\endgroup$ – user61527 May 31 '14 at 23:38
  • $\begingroup$ The chain rule for (3xy+1)^2 and do I move the 2x^3 over once I get the derivative, which is 6x^2. $\endgroup$ – user154479 May 31 '14 at 23:39
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    $\begingroup$ @user154479 Please edit your original question to include that. $\endgroup$ – user61527 May 31 '14 at 23:41
  • $\begingroup$ The answer is (-3y^2x-y+x^2)/(x(3yx+1)), I am have trouble solving this problem. $\endgroup$ – user154479 May 31 '14 at 23:42

First, it may be helpful to look at patrickJMT's video on implicit differentiation.

Being careful to use the chain rule when needed, take the derivative of both sides with respect to $x$,

$$6x^2 = 2(3xy+1)\left(3y + 3x \frac{dy}{dx}\right).$$

Expand the RHS to obtain $$6x^2 = 18x^2 y\frac{dy}{dx} + 6 x\frac{dy}{dx} + 18xy^2 + 6y.$$ Rearranging, $$6x^2 - 18xy^2 - 6y = \frac{dy}{dx}(18x^2y + 6x).$$ Which gives

$$\frac{dy}{dx} = \frac{6x^2 - 18xy^2 - 6y}{18x^2y + 6x}.$$

Of course, you could clean this up a bit if desired.

  • $\begingroup$ Ok, I understand what you have done, would the next step be to move the 6x^2 to the right or everything to the left except for the 3 dy/dx? $\endgroup$ – user154479 May 31 '14 at 23:52
  • $\begingroup$ @user154479 I updated my answer to include more details. If you still have any questions, just let me know. $\endgroup$ – Gamma Function Jun 1 '14 at 0:00

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