# Implicit differentation

• February 25th 2009, 02:20 PM
mathamatics112
Implicit 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
• February 25th 2009, 02:35 PM
skeeter
Quote:

Originally Posted by mathamatics112
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

show what you tried ...
• February 25th 2009, 02:39 PM
Amanda H
Product rule does come into it with the first part of the equation but you need to apply the chain rule as well.

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

first of all when you differentiate y it should become $\frac{dy}{dx}$
So applying the chain rule to $y^2$ you get:

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

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

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

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

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

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

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

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

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

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

Differentiating y can be confusing so I can try to go over it in more detail if you want.
• February 25th 2009, 02:40 PM
mathamatics112
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.
• February 25th 2009, 02:43 PM
Soroban
Hello, mathamatics112!

Quote:

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

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

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

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

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

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

Quote:

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

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

Then: . $\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
$

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

• February 25th 2009, 02:49 PM
mathamatics112
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 :http://mathhelpforum.com/calculus/75755-implicit-differentation.html#post272669\" rel=\"nofollow\">
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

It's easiest to use Implicit Differentiation if you realise that since both sides of the equation are equal, so are their derivatives.

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

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

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

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

So, using these pieces of information we have

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

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

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

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

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

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

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