If Z = f(u, v), u = 𝑥^2−𝑦,𝑣=𝑥−𝑦^2, find Z𝑥𝑥 in terms of derivatives of f.

Question

If Z = f(u, v), u = 𝑥^2−𝑦,𝑣=𝑥−𝑦^2, find Z𝑥𝑥 in terms of derivatives of f.

No clue what to do to solve. I try looking for a video but no luck.Show me in a way so I understand how to solve it on my own.

Answer 1

just apply the chain rule.

Zx = ∂f/∂x = ∂f/∂u * ∂u/∂x + ∂f/∂v * ∂v/∂x = 2x ∂f/∂u + 1 ∂f/∂v
now use that to find
Zxx = ∂Zx/∂x

Answer 2

this article might be helpful.

https://math.stackexchange.com/questions/2312943/multivariate-chain-rule-and-second-order-partials

Answer 3

To find Z𝑥𝑥 in terms of derivatives of f, we'll need to differentiate Z with respect to x twice.

Given Z = f(u, v), where u = 𝑥^2−𝑦 and v = 𝑥−𝑦^2, we need to apply the chain rule to differentiate Z with respect to x.

First, let's find the partial derivatives of Z with respect to u and v:

∂Z/∂u = ∂f/∂u
∂Z/∂v = ∂f/∂v

Next, we'll differentiate u and v with respect to x:

du/dx = 2x
dv/dx = 1

Using the chain rule, we can express ∂Z/∂x in terms of ∂Z/∂u, ∂Z/∂v, du/dx, and dv/dx:

∂Z/∂x = (∂Z/∂u) * (du/dx) + (∂Z/∂v) * (dv/dx)

Now, let's differentiate ∂Z/∂x with respect to x. Since (∂Z/∂u) depends on u, and (∂Z/∂v) depends on v, we'll need to differentiate (∂Z/∂u) and (∂Z/∂v) with respect to u and v, respectively.

To differentiate (∂Z/∂u) with respect to u, we can use the chain rule again:

∂²Z/∂u² = (∂(∂Z/∂u)/∂u) * (du/dx)

Similarly, to differentiate (∂Z/∂v) with respect to v:

∂²Z/∂v² = (∂(∂Z/∂v)/∂v) * (dv/dx)

Finally, to find Z𝑥𝑥, we can sum up these partial derivatives:

Z𝑥𝑥 = ∂²Z/∂u² + ∂²Z/∂v²

So, by applying the chain rule twice, we can find Z𝑥𝑥 in terms of the derivatives of f.