# Thread: Find y'' by implicit differentiation help!

1. ## Find y'' by implicit differentiation help!

x^3 + y^3 =6xy

No need for d/dx just dy/dx for when you find the derivative of y only.

My steps:

y'= 3x^2+3y^2(dy/dx)= 6y+6x(dy/dx)

dy/dy = (3x^2-6y) /(6x-3y^2)
= (x^2-2y) / (2x-y^2)

y'' = [ (2x- 2(dy/dx)) (2x-y^2) ] - [(x^2-2y)(2-2y(dy/dx))]
= [ 2x-2(x^2-2y/2x-y^2) ] - [(x^2-2y)(2-2y(x^2-2y/2x-y^2)) ] / (2x-y^2)^2

If your having trouble understanding, when i find the derivative of y, its the derivative and dy/dx, where i replace dy/dx in y'' with the dy/dx from y' .

I dont know how to actually put symbols i cant find any button to place actually division signs and such so i clear this up a bit

So then i simplified the complex fraction by multiplying by 2x-y^2

[ 2x(2x-y^2)-2(x^2-2y) ] - (x^2-2y)(2x-y^2) (2(2x-y^2)) (2y(x^2-2y))
4x^2-2y^2 -2x^2-4y - 4y(x^2-2y)^2(2x-y^2)^2
2x^2-2xy^2-4y -4y(x^2-2y)^2(2x-y^2)^2 / (2x-y^2)^2

Is any of this right, that final shown above, not final but mid way, seems way too complicated and much work for such a simple question. All that factoring and such seems alot.

If clarification is needed dont hesitate

2. ## Re: Find y'' by implicit differentiation help!

OK, for starters, it helps if you don't switch between Leibnitz and Newtonian notation. Stick to one or the other. I tend to prefer Leibnitz.

Second, you need to understand that \displaystyle \begin{align*} \frac{d}{dx} \end{align*} is an OPERATOR meaning "take the derivative with respect to x", while \displaystyle \begin{align*} \frac{dy}{dx} \end{align*} is a function, namely the derivative of y with respect to x. It can also be written as y'(x) if you prefer Newtonian notation.

The basic idea is that y is a function of x, so if you have a function of y, you can differentiate it with respect to x.

\displaystyle \begin{align*} x^3 + y^3 &= 6\,x\,y \\ \frac{d}{dx} \left( x^3 + y^3 \right) &= \frac{d}{dx} \left( 6\,x\,y \right) \\ \frac{d}{dx} \left( x^3 \right) + \frac{d}{dx} \left( y^3 \right) &= 6 \left[ x\,\frac{d}{dx} \left( y \right) + y\,\frac{d}{dx} \left( x \right) \right] \\ 3x^2 + \frac{d}{dy} \left( y^3 \right) \frac{dy}{dx} &= 6 \left[ x\,\frac{dy}{dx} + y \left( 1 \right) \right] \\ 3x^2 + 3y^2\,\frac{dy}{dx} &= 6x\,\frac{dy}{dx} + 6y \\ \left( 3y^2 - 6x \right) \frac{dy}{dx} &= 6y - 3x^2 \\ 3 \left( y^2 - 2x \right) \frac{dy}{dx} &= 3 \left( 2y - x^2 \right) \\ \frac{dy}{dx} &= \frac{2y - x^2}{y^2 - 2x} \end{align*}

Now take the derivative of both sides again to find the second derivative.