Expert: Alon Mandes - 10/11/2009

Question
Use implicit differentiation to find the second derivative of:
2xy^2=4  (nothing is in parenthesis)

I understand how to do simple differentiation, but I get confused with this problem since there is more than one variable.

Answer
Ok Tiffany, let's review %26 sharpen some of the basic facts concerning implicit derivation :
1st of all, for any function y=f(x) , we derive it by performing the differentiate operator
on both sides of the equation. "Differentiate operator is simply d/dx" . Say :
d/dx {y=f(x)}
d/dx {y} = d/dx {f(x)}
y'=f'(x).
Now, if the function is not defined as y=f(x) but in an implicit way, such as f(x,y)=0 .
In this case, we perform the differentiate operator, only this time , we treat y as a dependent
variable of x, d/dx {y} = y'. Further more, the derivative of the form y^2 will be :
d/dx {y^2} = 2yy' .
Lets get back to our case:
2xy^2=4
d/dx {2xy^2} = d/dx {4}
Here we have to derive a product of 2 functions. According to the rule : [fg]'=f'*g+g'*f. In
our case f=2x %26 g=y^2, therefore :
d/dx {2x} * y^2 + 2x * d/dx {y^2} = 0
2y^2+4xyy'=0
y'=-(y^2)/(2xy) .

Now, lets derive second time : meaning, let's perform the differentiate operator on the
equation : 2y^2+4xyy'=0 ,
d/dx {2y^2+4xyy'=0}
d/dx {2y^2+4xyy'} = d/dx {0}
d/dx {2y^2} + d/dx {4xyy'} = 0
Here we have to derive a product of 2 functions. According to the rule : [fg]'=f'*g+g'*f. In
our case f=4xy %26 g=y', therefore :
2yy' + d/dx {4xy} * y' + 4xy * d/dx {y'} = 0
%26 another time , we have to derive a product of 2 functions. According to the rule : [fg]'=f'*g+g'*f. In our case f=4x %26 g=y, thus,
2yy' + d/dx {4x} * y * y' + 4x * d/dx {y} * y' + 4xyy'' = 0
2yy'+4yy'+4x(y')^2+4xyy''=0
6yy'+4x(y')^2+4xyy''=0
3yy'+2x(y')^2+2xyy''=0
y''=-[3yy'+2x(y')^2]/[2xy].
We already calculated y', so we can substitute it's value {"y'=-(y^2)/(2xy)"}:
y''=-[ -3y*y^2/(2xy) + 2x{(y^2)^2/(2xy}^4 ] / [2xy].
y''=-[ -3y^3/(2xy) + 2xy^4/(2xy}^4 ] / [2xy].
y''=-[ -3y^3/(2xy) + 1/(8x^3) ] / [2xy].

Alon.