Solving Chain Rule Problems: Methods and Solutions

Chain Rule Problem Solution

To begin, we must first recall that the total differential of z equals the partial
derivative of z in the x direction multiplied by dx plus the partial derivative of z in the
y direction multiplied by dy.
a) dz = z_x dx + z_y dy
Now, looking at our formula for z, we see that the partial derivative in the x direction
is 2x, and the partial derivative in the y direction is 2y.
dz = 2x dx+2y dy v
Next, we will solve for this partial derivative with respect to x using the chain rule.

The chain rule states that whenever we compute a derivative, we can break down
the calculation into smaller steps by working on one term at a time and keeping track
of the relationships between them. So let's create a dependency graph. The equation
below shows how the variables in my model are interrelated. At the top, we have z.

z
is a function of x and y, but x is itself a function of both u and v. And y is also a
function of u and v; that is, z depends on x and y. And x and y each jointly depend on
u and v-that is, none of these variables can be considered in isolation from any
others because they all interact with one another.
[Graph]
So, it's a little bit complicated, the relationships here. Now, what the chain rule says
is that. if we take a partial derivative of z with respect to u, we have to go through our
dependency graph every way that we can get from z to u.

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We get a term in our
summation for each one of those. For example, z goes to x goes to u. This means
that we have partial z, partial x, partial x and partial u. Alternatively, we can also go z
goes to y goes to u, which gives us partial z, partial y and partial y.
b) δz/δu=δz/δx δx/δu+δz/δy δy/δu
Now we will compute some of the partial derivatives. Partial z partial x is 2x, which
we computed earlier. Partial x partial u is 2u; this result makes sense because x was
defined as u squared minus v squared, so if we take a partial derivative in the u
direction, we get 2u by definition. Similarly, partial y partial u is v again because y
was defined as uv; therefore, we take a partial derivative in the u direction to get v.
δz/δu=2x*2u+2yv
So in fact, this is 4ux plus 2vy:
δz/δu=4ux+2vy
And that's our partial derivative. So, notice that x is a function of u and v. So, we
could substitute an expression for x using its formula V. for u and v. but that's not really
necessary. What's interesting about these problems is how the differentials depend
on one another. And we re perfectly happy with an answer that has mixed variables
like this.
Now let us examine the problem under a new light. We will use a different method of
differentiation to arrive at the solution. Use total differentials, because the chain rule
can be more intuitively understood when using them. We will compute dz in this case
by differentiating both sides of the equation with respect to z.
dz=2xdx+2ydy
Now we want to use the fact that x is itself a function of u and v. So that's what we
need to do now. Now, d dx equals 2u du minus 2v dv.
dx = 2u du – 2v dv
And dy, sO remember, y was. uv so taking d of uv, we get v du plus u dv.
dy = v du + u dv
We now have all the partial derivatives, and we can move on to solving for z. We
only need to substitute our formulas for dx into the formula for dz and divide by 2.
dz = 2x(2u du – 2v dv)
So that was this term. And now we have plus 2y. V du plus u dv.
dz = 2x(2u du – 2v dv) + 2y(v du + u dv) +
It's just substitution, so now we just expand everything out using – just expanding
these out. So if we collect all the things involving du, we have 4xu plus 2yv and this
whole quantity times du. And then if we collect terms in dv, we have 2yu. So that's
coming from here. And then we have a minus 4xv.
dz= 2x 2u du – 2v dv
) + 2y udu + u dv
) = Abu + 2vy
) du + 2ju – 4xv
And now we know that the partial derivative, written as dz/du or dz/dv, is this
coefficient. So if we write the total differential as (dz/du)+(dz/dv), we can see that it
equals the partial derivative of z with respect to u plus the partial derivative of z with
respect to v.
)dv
dz=δz/δu du+δz/δv dv
Let us consider an expression for parcels of the type z partial u. If we let this
coefficient equal 4xu plus 2, one of these terms is an x, and one is a v, then we can
find that partial z partial u is equal to 4xu plus 2vy.
δz/δu=4(xu+2vy)
Now, just as a sanity check to make sure we've covered this topic well, let's go back
to the middle of the board and see that we got the same thing. So 4xu plus 2vy is
what we concluded for partial z partial u. And now let's go over the two different
methods and compare them.
If you are in a rush, use the dependency graph along with tracing paths from z to u.
By multiplying all of the partial derivatives that correspond to each edge and adding
them up, we get an expression. If you have more time, then use the method of total
differentials. This requires some basic calculus and some algebraic manipulations.
As you saw, this method involved computing a greater number of derivatives that
were not used in the final answer. However, believe that this method is more
conceptually straightforward than the first.