Given that ; $$ z=5 x^{3}+3 x^{2} y+4 y^{3} $$ find $$ \frac{d z}{d x}, \frac{d z}{d y}, d x^{2} \frac{d^{2} z}{d x^{2}}, \frac{d^{2} z}{d y^{2}}, \frac{d^{2} z}{d y d x} $$ and $$ \frac{d^{2} z}{d x d y} $$

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Alright, I need to find the first and second partial derivatives of the function $$ z = 5x^{3} + 3x^{2}y + 4y^{3} $$ with respect to $$ x $$ and $$ y $$, as well as the mixed partial derivatives. Let's break this down step by step.

First, let's recall that partial derivatives are taken with respect to one variable while keeping the others constant. So, $$ \frac{d z}{d x} $$ means differentiating $$ z $$ with respect to $$ x $$ while treating $$ y $$ as a constant, and similarly for $$ \frac{d z}{d y} $$.

Starting with $$ \frac{d z}{d x} $$:
$$
\frac{d z}{d x} = \frac{d}{d x}(5x^{3}) + \frac{d}{d x}(3x^{2}y) + \frac{d}{d x}(4y^{3})
$$
Calculating each term:
- $$ \frac{d}{d x}(5x^{3}) = 15x^{2} $$
- $$ \frac{d}{d x}(3x^{2}y) = 6xy $$ (since $$ y $$ is treated as a constant)
- $$ \frac{d}{d x}(4y^{3}) = 0 $$ (since $$ y $$ is treated as a constant)

So,
$$
\frac{d z}{d x} = 15x^{2} + 6xy
$$

Next, $$ \frac{d z}{d y} $$:
$$
\frac{d z}{d y} = \frac{d}{d y}(5x^{3}) + \frac{d}{d y}(3x^{2}y) + \frac{d}{d y}(4y^{3})
$$
Calculating each term:
- $$ \frac{d}{d y}(5x^{3}) = 0 $$ (since $$ x $$ is treated as a constant)
- $$ \frac{d}{d y}(3x^{2}y) = 3x^{2} $$ (since $$ x $$ is treated as a constant)
- $$ \frac{d}{d y}(4y^{3}) = 12y^{2} $$

So,
$$
\frac{d z}{d y} = 3x^{2} + 12y^{2}
$$

Now, let's find the second partial derivatives. Starting with $$ \frac{d^{2} z}{d x^{2}} $$:
$$
\frac{d^{2} z}{d x^{2}} = \frac{d}{d x}\left(15x^{2} + 6xy\right) = 30x + 6y
$$

Next, $$ \frac{d^{2} z}{d y^{2}} $$:
$$
\frac{d^{2} z}{d y^{2}} = \frac{d}{d y}\left(3x^{2} + 12y^{2}\right) = 0 + 24y = 24y
$$

Now, the mixed partial derivatives. First, $$ \frac{d^{2} z}{d y d x} $$:
$$
\frac{d^{2} z}{d y d x} = \frac{d}{d y}\left(15x^{2} + 6xy\right) = 6x
$$

Similarly, $$ \frac{d^{2} z}{d x d y} $$:
$$
\frac{d^{2} z}{d x d y} = \frac{d}{d x}\left(3x^{2} + 12y^{2}\right) = 6x
$$

Wait a second, both $$ \frac{d^{2} z}{d y d x} $$ and $$ \frac{d^{2} z}{d x d y} $$ are equal to $$ 6x $$. This makes sense because for functions that are twice differentiable, the mixed partial derivatives are equal (Clairaut's Theorem).

Let me double-check the calculations to ensure there are no mistakes.

Reviewing $$ \frac{d z}{d x} $$:
$$
\frac{d z}{d x} = 15x^{2} + 6xy
$$
This seems correct.

Reviewing $$ \frac{d z}{d y} $$:
$$
\frac{d z}{d y} = 3x^{2} + 12y^{2}
$$
Also correct.

Second derivatives:
$$
\frac{d^{2} z}{d x^{2}} = 30x + 6y
$$
$$
\frac{d^{2} z}{d y^{2}} = 24y
$$
These are accurate.

Mixed partials:
$$
\frac{d^{2} z}{d y d x} = 6x
$$
$$
\frac{d^{2} z}{d x d y} = 6x
$$
Consistent and correct.

Everything checks out. The mixed partial derivatives are indeed equal, which is expected for this function.

$$ \frac{d z}{d x} = 15x^{2} + 6xy $$ $$ \frac{d z}{d y} = 3x^{2} + 12y^{2} $$ $$ \frac{d^{2} z}{d x^{2}} = 30x + 6y $$ $$ \frac{d^{2} z}{d y^{2}} = 24y $$ $$ \frac{d^{2} z}{d y d x} = 6x $$ $$ \frac{d^{2} z}{d x d y} = 6x $$ \]

Given that ; \( z=5 x^{3}+3 x^{2} y+4 y^{3} \) find \( \frac{d z}{d x}, \frac{d z}{d y}, d x^{2} \frac{d^{2} z}{d x^{2}}, \frac{d^{2} z}{d y^{2}}, \frac{d^{2} z}{d y d x} \) and \( \frac{d^{2} z}{d x d y} \)

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