https://file.aiquickdraw.com/mathbot/file/1730885143358u5lk2xql.jpg

Let's analyze the function given:

\[
f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}
\]

and calculate the following partial derivatives as requested.

### a) \(\frac{\partial f}{\partial z}\)

To find \(\frac{\partial f}{\partial z}\), we differentiate \(f(x, y, z)\) with respect to \(z\), treating \(x\) and \(y\) as constants:

\[
f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}
\]

Applying the product rule, we get:

\[
\frac{\partial f}{\partial z} = \cos(2x) \cdot \frac{\partial}{\partial z} \left(z \cdot e^{y \cdot z}\right)
\]

Expanding \(\frac{\partial}{\partial z} \left(z \cdot e^{y \cdot z}\right)\) with the product rule:

\[
\frac{\partial}{\partial z} \left(z \cdot e^{y \cdot z}\right) = e^{y \cdot z} + z \cdot y \cdot e^{y \cdot z} = e^{y \cdot z} (1 + y \cdot z)
\]

Thus,

\[
\frac{\partial f}{\partial z} = \cos(2x) \cdot e^{y \cdot z} \cdot (1 + y \cdot z)
\]

### b) \(\frac{\partial f}{\partial x}\)

To find \(\frac{\partial f}{\partial x}\), we differentiate \(f(x, y, z)\) with respect to \(x\), treating \(y\) and \(z\) as constants:

\[
f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}
\]

The derivative of \(\cos(2x)\) with respect to \(x\) is \(-2 \sin(2x)\), so:

\[
\frac{\partial f}{\partial x} = -2 \sin(2x) \cdot z \cdot e^{y \cdot z}
\]

### c) \(\frac{\partial^2 f}{\partial x \partial y}\)

To find \(\frac{\partial^2 f}{\partial x \partial y}\), we first find \(\frac{\partial f}{\partial y}\) and then differentiate it with respect to \(x\):

\[
f(x, y, z) = \cos(2x) \cdot z \cdot e^{y \cdot z}
\]

First, we compute \(\frac{\partial f}{\partial y}\):

\[
\frac{\partial f}{\partial y} = \cos(2x) \cdot z \cdot \frac{\partial}{\partial y} \left(e^{y \cdot z}\right) = \cos(2x) \cdot z^2 \cdot e^{y \cdot z}
\]

Now, we differentiate this with respect to \(x\):

\[
\frac{\partial^2 f}{\partial x \partial y} = -2 \sin(2x) \cdot z^2 \cdot e^{y \cdot z}
\]

### d) \(\frac{\partial^2 f}{\partial y^2}\)

To find \(\frac{\partial^2 f}{\partial y^2}\), we first find \(\frac{\partial f}{\partial y}\) and then differentiate it again with respect to \(y\):

We already have:

\[
\frac{\partial f}{\partial y} = \cos(2x) \cdot z^2 \cdot e^{y \cdot z}
\]

Now, differentiate this with respect to \(y\):

\[
\frac{\partial^2 f}{\partial y^2} = \cos(2x) \cdot z^3 \cdot e^{y \cdot z}
\]

### Summary of the results:

1. \(\frac{\partial f}{\partial z} = \cos(2x) \cdot e^{y \cdot z} \cdot (1 + y \cdot z)\)
2. \(\frac{\partial f}{\partial x} = -2 \sin(2x) \cdot z \cdot e^{y \cdot z}\)
3. \(\frac{\partial^2 f}{\partial x \partial y} = -2 \sin(2x) \cdot z^2 \cdot e^{y \cdot z}\)
4. \(\frac{\partial^2 f}{\partial y^2} = \cos(2x) \cdot z^3 \cdot e^{y \cdot z}\)

Would you like any further explanation or details?

Here are some follow-up questions that expand on this topic:

1. How would the partial derivatives change if \(f(x, y, z)\) had a different exponent in the exponential term?
2. What would be the result of \(\frac{\partial^2 f}{\partial z \partial x}\)?
3. Can we compute the third derivative \(\frac{\partial^3 f}{\partial x \partial y \partial z}\) for this function?
4. How does the cosine function affect the partial derivatives compared to if it was a simple constant?
5. How would the answer change if \(f(x, y, z) = \sin(2x) \cdot z \cdot e^{y \cdot z}\)?

**Tip:** Always remember to apply the chain rule carefully when dealing with exponential functions and trigonometric expressions in partial derivatives.

Laske seuraavat osittaisderivaatit annetuista funktioista.

Learn how to compute partial derivatives of a multivariable function f(x, y, z) = cos(2x) · z · e^(y · z) with respect to variables z, x, and y. This solution involves using product and chain rules for exponential and trigonometric functions, suitable for university-level calculus students.

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