implicit differentiation: trying to find 2nd deriv of dy/dx=1/(xy)

Well, the second derivative is the derivative of the derivative, so I'd start by finding the first derivative and then go from there. Have you found the first derivative? If so, what is it? What happens if you try to take the derivative of that? As a hint, to find the first derivative, you'll want to use the Quotient Rule and the Product Rule.
 
ksdhart2 said:
Well, the second derivative is the derivative of the derivative, so I'd start by finding the first derivative and then go from there. Have you found the first derivative? If so, what is it? What happens if you try to take the derivative of that? As a hint, to find the first derivative, you'll want to use the Quotient Rule and the Product Rule.
Click to expand...


1/(xy) is the first derivative, I'm just trying to find the derivative of that equation. Whatever my answer equation is needs to be where I can plug in an x and an y value. My problem is I end up without a y variable.
dy=dx/xy
dyy=dx/x
1=-x^-2
is what I'm getting. I think I'm messing up the implicit differentiation somewhere?
 
Given that the first derivative is dy/dx= 1/xy then I would start by writing it as ydy/dx= 1/x and differentiate both sides with respect to x. The product rule, on the left, gives (dy/dx)(dy/dx}+ y(d^2y/dx^2) while the derivative of 1/x= x^-1 is -1/x^2. So (dy/dx)^2+ xy(d^2y/dx^2)= -y^2. We know that dy/dx= 1/xy so that becomes (1/xy)^2+ (1/xy)(d^2y/dx^2)= -1/x^2. Multiply on both sides by x^2y^2: 1+ xy(d^2y/dx^2)= -y^2. Solving for (d^2y/dx^2)= -(y^2- 1)/xy.
 
hellovermont said:
1/(xy) is the first derivative, I'm just trying to find the derivative of that equation. Whatever my answer equation is needs to be where I can plug in an x and an y value. My problem is I end up without a y variable.
dy=dx/xy
dyy=dx/x
1=-x^-2
is what I'm getting. I think I'm messing up the implicit differentiation somewhere?
Click to expand...
I'm not at all clear on what you're doing here...? You have:

. . . . .dydx = 1xy\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{1}{xy}

You need to find:

. . . . .d2ydx2\displaystyle \dfrac{d^2y}{dx^2}

So you need to apply the Quotient Rule (if you're not doing the multiply-through-by-y suggestion earlier in this thread). So you get:

. . . . .d2ydx2 = ((xy) ddx(1) − (1) ddx(xy))(xy)2\displaystyle \dfrac{d^2y}{dx^2}\, =\, \dfrac{\left((xy)\,\dfrac{d}{dx}(1)\, -\, (1)\, \dfrac{d}{dx}(xy)\right)}{(xy)^2}

...and,... then what? Either method you choose, please show all of your steps. Thank you! ;)
 
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