Derivative of a right triangle's area with respect to its height I want to find the first derivative of the area of a right triangle as its non-hypotenuse sides change as a function of a third variable.  I try it two different ways and get two different answers. The question has arisen as part of calculating producer surplus (the area beneath a horizontal price curve, but above a marginal cost curve). The problem is:
Calculate
$\frac{\partial Area}{\partial z}$
where x and y, both functions of z, are sides of the triangle (let's let x be the base, and y be the height)
$Area = \frac{1}{2}x(z)y(z)$
Approach 1 -- Geometric:
The change in the area of the triangle as z changes is the box formed by $x\frac{\partial y}{\partial z}$ plus the additional tiny triangle $1/2 \frac{\partial x}{\partial z} \frac{\partial y}{\partial z}$
Solution 1:
$\frac{\partial Area}{\partial z} = x\frac{\partial y}{\partial z} +1/2 \frac{\partial x}{\partial z} \frac{\partial y}{\partial z} $
Approach 2 -- Algebraic:
Take the derivative of area with respect to z. Using product rule yields:
Solution 2:
$\frac{\partial Area}{\partial z} = 1/2(\frac{\partial y}{\partial z}x + y\frac{\partial x}{\partial z})$
if, as I think must be the case $\frac{\partial y}{\partial z}x =y\frac{\partial x}{\partial z} $, then this reduces to
$\frac{\partial Area}{\partial z} = \frac{\partial y}{\partial z}x$
which is just the "box" part of the above geometric approach.
What am I missing here? My instincts tell me the first one is correct, as the second seems wrong when thinking about it geometrically, but I can't tell how I screwed up the calculus.
Edit: heres a diagram showing where my geometric intuition is coming from: https://imgur.com/9m1vrFG
Edit 2: I agree the second bit of solution 2 is incorrect
 A: The chain rule immediately gives $\frac{d{\rm Area}}{dz} = \frac12\bigl( x'(z)y(z) + x(z)y'(z) \bigr)$ as the correct answer, which is the first formula under Solution 2. There's no reason to think that $x'(z)y(z) = x(z)y'(z)$ in general. The first approach calculated the change in area near the base but forgot the change in area near the vertical side; in addition, the term $x'(z)y'(z)$ shouldn't be there at all—that arises from conflating $\Delta x$, the change in $x$, with $x'(z)$, the rate of change in $x$.
A: The geometric approach:
Easier to think of the triangle as half the rectangle rather than working with the triangle directly.

The area of the rectangle is $xy$
If we give x and y a little nudge, the area of the perturbed rectangle is $A = xy + y dx + x dy + dx\ dy$
But if $dx, dy$ are both small, then $dx\ dy$ will be very small and can be ignored.
$dA = x dy + y dx$  Which is the same thing the algebraic approach gives.
Now, if we are working with triangles then we can divide everything by 2.
$dA =  \frac 12 (x\ dy + y\ dx)$
