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
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.
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.
Hello, mathamatics112!
$\displaystyle 3xy^2 + y^3 \:=\: 8$
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} $
$\displaystyle \frac{x^2}{16} + \frac{3y^2}{13} \:=\: 1$
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
$
. . . . $\displaystyle \frac{dy}{dx} \;=\;\frac{-\frac{1}{8}x}{\frac{6}{13}y} \quad\Rightarrow\quad \frac{dy}{dx} \;=\;-\frac{13x}{48y}$
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.