Need help using implicit differentiation on equation:

3xy^2 + y^3 = 8

x^2/16 + 3y^2/13 = 1

I just do not know how to start it at all :( I tried using product rule and it didn't work

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- Feb 25th 2009, 02:20 PMmathamatics112Implicit differentation
Need help using implicit differentiation on equation:

3xy^2 + y^3 = 8

x^2/16 + 3y^2/13 = 1

I just do not know how to start it at all :( I tried using product rule and it didn't work - Feb 25th 2009, 02:35 PMskeeter
- Feb 25th 2009, 02:39 PMAmanda H
Product rule does come into it with the first part of the equation but you need to apply the chain rule as well.

$\displaystyle 3xy^2 + y^3 = 8 $

first of all when you differentiate y it should become $\displaystyle \frac{dy}{dx} $

So applying the chain rule to $\displaystyle y^2 $ you get:

$\displaystyle 2y\frac{dy}{dx} $ you deal with $\displaystyle y^3 $ in a similar way.

Applying all this with the product rule the whole function should become:

$\displaystyle 3*y^2 + 3x*2y\frac{dy}{dx} + 3y^2*\frac{dy}{dx} = 0 $

You now need to rearrange to get $\displaystyle \frac{dy}{dx} = ... $

Put the $\displaystyle 3y^2 $ on the other side

$\displaystyle 6xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} = -3y^2 $

Take out $\displaystyle \frac{dy}{dx} $ as a common factor

$\displaystyle \frac{dy}{dx} (6xy + 3y^2) = -3y^2 $

Then divide to leave $\displaystyle \frac{dy}{dx} $ on its own

$\displaystyle \frac{dy}{dx} = \frac{-3y^2}{6xy + 3y^2} $

Differentiating y can be confusing so I can try to go over it in more detail if you want. - Feb 25th 2009, 02:40 PMmathamatics112
3xy^2 + y^3 = 8

now product rule...

(3)(y^2)dx/dy + (3x)(2y) dx/dy +3y^2 dx/dy = 0

I have all the dx/dy on one side... now what? This is where I went wrong and I do know know where...

Oh ok thanks alot, but when do you know to add the dy/dx beside one of the terms? I think that is where I am going wrong. - Feb 25th 2009, 02:43 PMSoroban
Hello, mathamatics112!

Quote:

$\displaystyle 3xy^2 + y^3 \:=\: 8$

We have: .$\displaystyle 3x\cdot y^2 + y^3 \:=\:8$

Then: .$\displaystyle 3x\cdot2y\,\frac{dy}{dx} + 3y^2 + 3y^2\,\frac{dy}{dx} \:=\:0$

. . $\displaystyle 6xy\,\frac{dy}{dx} + 3y^2\,\frac{dy}{dx} \:=\:-3y^2$

. . $\displaystyle 3y(2x+y)\,\frac{dy}{dx} \:=\:-3y^2$

. . $\displaystyle \frac{dy}{dx} \:=\:\frac{-3y^2}{3y(2x+y)} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{y}{2x+y} $

Quote:

$\displaystyle \frac{x^2}{16} + \frac{3y^2}{13} \:=\: 1$

We have: .$\displaystyle \frac{1}{16}x^2 + \frac{3}{13}y^2 \:=\:1$

Then: .$\displaystyle \frac{1}{8}x + \frac{6}{13}y\,\frac{dy}{dx} \:=\:0 \quad\Rightarrow\quad \frac{6}{13}y\,\frac{dy}{dx} \:=\:-\frac{1}{8}x

$

. . . . $\displaystyle \frac{dy}{dx} \;=\;\frac{-\frac{1}{8}x}{\frac{6}{13}y} \quad\Rightarrow\quad \frac{dy}{dx} \;=\;-\frac{13x}{48y}$

- Feb 25th 2009, 02:49 PMmathamatics112
Thank you very much, but I was wondering, When do you add the dy/dx? I put it in the wrong areas, which is why I am getting these answers wrong :(

- Feb 25th 2009, 02:55 PMAmanda H
You need to put in $\displaystyle \frac{dy}{dx} $ whenever you differentiate $\displaystyle y, y^2, y^3, etc $ along with the use of the chain rule if necessary.

- Feb 25th 2009, 03:27 PMProve It
It's easiest to use Implicit Differentiation if you realise that since both sides of the equation are equal, so are their derivatives.

So $\displaystyle \frac{d}{dx}(3xy^2 + y^3) = \frac{d}{dx}(8)$

Notice on the LHS you have a sum, and the derivative of a sum is the same as the sum of the derivatives. On the right you have a constant, and the derivative of a constant is 0.

So we have

$\displaystyle \frac{d}{dx}(3xy^2) + \frac{d}{dx}(y^3) = 0$.

The first part is a product, so we need to use the product rule. Since y is a function of x, we can find a derivative with respect to x, in terms of y. We do this using the chain rule.

Notice that $\displaystyle \frac{dy}{dx}\cdot\frac{d}{dy} = \frac{d}{dx}$. Also note that we are trying to FIND $\displaystyle \frac{dy}{dx}$.

So, using these pieces of information we have

$\displaystyle 3\left[x\frac{d}{dx}(y^2) + y^2\frac{d}{dx}(x)\right] + \frac{dy}{dx}\cdot\frac{d}{dy}(y^3) = 0$

$\displaystyle 3\left[x\frac{dy}{dx}\cdot\frac{d}{dy}(y^2) + y^2\right] + 3y^2\frac{dy}{dx} = 0$

$\displaystyle 3\left(2xy\frac{dy}{dx} + y^2\right) + 3y^2\frac{dy}{dx} = 0$

$\displaystyle 6xy\frac{dy}{dx} + 3y^2 + 3y^2\frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx}(6xy + 3y^2) = -3y^2$

$\displaystyle \frac{dy}{dx} = -\frac{3y^2}{6xy + 3y^2}$.

Hope that helped. Have a go of the second one yourself.