Nicole M, Maddie B, Jordan, Shelly, Maddie G, Emma, Nora
Taking the 2nd Implicit Derivative
Taking the second implicit derivatives is a little more complicated than simply taking the first. To take the second implicit derivative involves taking the derivative much like the first method, although the first derivative must be subbed in for every (dy/dx) present in the second derivative. We'll use the first derivative we found during the "Taking the First Implicit Derivative" example:
​
=
​
g'(x) = -2y - 2x(dy/dx) h'(x) = 12y (dy/dx) + 2x
​
d y = (-2y - 2x(dy/dx))(4y + x ) - (12y (dy/dx) + 2x)(1 - 2xy)
​
​
d y = -8y - 2x y - 8xy (dy/dx) - 2x (dy/dx) - (12y (dy/dx) - 24xy (dy/dx) + 2x - 4x y)
​
​
d y = -8y - 2x y - 8xy (dy/dx) - 2x (dy/dx) - 12y (dy/dx) + 24xy (dy/dx) - 2x + 4x y
​
​
d y = -8y + 2x y + 16xy (dy/dx) - 2x (dy/dx) - 12y (dy/dx) - 2x
​
​
(dy/dx)
1 - 2xy
4y + x
3
2
2
(4y + x )
(4y + x )
(4y + x )
(4y + x )
3
3
3
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
4
4
4
dx
dx
dx
dx
2
2
2
2
-8y + 2x y + 16xy - 2x - 12y - 2x
1 - 2xy
1 - 2xy
1 - 2xy
4y + x
4y + x
4y + x
_______________________________________________________
(4y + x )
=
d y
dx
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
4
-8y + 2x y + - - - 2x
16xy - 32x y
2x - 4x y
12y - 24xy
4y + x
4y + x
4y + x
__________________________________________________________
=
(4y + x )
d y
dx
4
4
4
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
-8y + 2x y + - 2x
16xy - 32x y - 2x + 4x y - 12y + 24xy
4y + x
_____________________________________________________
(4y + x )
=
d y
dx
4
4
4
3
3
3
3
3
2
2
2
2
2
2
2
2
With all of the (dy/dx) replaced with the answer from the previous exercise, and everything multiplied and added together as much as possible, this is the answer we get. The answer is pretty messy, but this is typically as far as we have to go as much as implicit derivatives are concerned.