Given the function
f (y, z) = (sin y)ez3−z−1
+ eyz
(a) Determine fz
(b) Determine fy
(c) Determine fyz

Given the function \( f(y, z) = (\sin y) e^{z^3 - z - 1} + e^{yz} \), we will compute each of the partial derivatives step by step.

### (a) Determine \( f_z \) (the partial derivative with respect to \( z \)):

We apply the rules of differentiation, treating \( y \) as a constant.

#### First term: 
\[ (\sin y) e^{z^3 - z - 1} \]

The derivative of \( e^{z^3 - z - 1} \) with respect to \( z \) is:
\[ \frac{d}{dz}\left(e^{z^3 - z - 1}\right) = e^{z^3 - z - 1} \cdot \frac{d}{dz}(z^3 - z - 1) = e^{z^3 - z - 1} \cdot (3z^2 - 1) \]

Thus, the derivative of the first term is:
\[ (\sin y) \cdot e^{z^3 - z - 1} \cdot (3z^2 - 1) \]

#### Second term:
\[ e^{yz} \]

The derivative of \( e^{yz} \) with respect to \( z \) is:
\[ \frac{d}{dz}(e^{yz}) = y e^{yz} \]

So the partial derivative \( f_z \) is:
\[
f_z = (\sin y) e^{z^3 - z - 1} (3z^2 - 1) + y e^{yz}
\]

### (b) Determine \( f_y \) (the partial derivative with respect to \( y \)):

We now differentiate with respect to \( y \), treating \( z \) as constant.

#### First term:
\[ (\sin y) e^{z^3 - z - 1} \]

The derivative of \( \sin y \) is \( \cos y \). So, the derivative of the first term is:
\[ (\cos y) e^{z^3 - z - 1} \]

#### Second term:
\[ e^{yz} \]

The derivative of \( e^{yz} \) with respect to \( y \) is:
\[ \frac{d}{dy}(e^{yz}) = z e^{yz} \]

So the partial derivative \( f_y \) is:
\[
f_y = (\cos y) e^{z^3 - z - 1} + z e^{yz}
\]

### (c) Determine \( f_{yz} \) (the mixed partial derivative with respect to \( y \) and \( z \)):

Now, we differentiate \( f_y \) with respect to \( z \).

From part (b), we know that:
\[ f_y = (\cos y) e^{z^3 - z - 1} + z e^{yz} \]

#### First term:
\[ (\cos y) e^{z^3 - z - 1} \]

The derivative with respect to \( z \) is:
\[
(\cos y) \cdot e^{z^3 - z - 1} \cdot (3z^2 - 1)
\]

#### Second term:
\[ z e^{yz} \]

The derivative with respect to \( z \) is:
\[
\frac{d}{dz}(z e^{yz}) = e^{yz} + z y e^{yz}
\]

So the mixed partial derivative \( f_{yz} \) is:
\[
f_{yz} = (\cos y) e^{z^3 - z - 1} (3z^2 - 1) + e^{yz} + z y e^{yz}
\]

### Summary of Results:
(a) \( f_z = (\sin y) e^{z^3 - z - 1} (3z^2 - 1) + y e^{yz} \)

(b) \( f_y = (\cos y) e^{z^3 - z - 1} + z e^{yz} \)

(c) \( f_{yz} = (\cos y) e^{z^3 - z - 1} (3z^2 - 1) + e^{yz} + z y e^{yz} \)

Would you like more details or have any questions?

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Here are 5 related questions to expand on this topic:
1. What is the second partial derivative \( f_{zz} \) of the function?
2. How does the chain rule apply when computing partial derivatives?
3. Can you find the critical points of the function \( f(y, z) \)?
4. How would the function change if the exponent in \( e^{yz} \) were \( y^2 z^2 \)?
5. What happens if \( y = 0 \) or \( z = 0 \)? How does this simplify \( f(y, z) \)?

**Tip:** When differentiating exponentials with a variable in the exponent, remember to apply the chain rule carefully!

Given the function f(y, z) = (sin y)e^{z^3 - z - 1} + e^{yz}, (a) Determine f_z (b) Determine f_y (c) Determine f_{yz}

Explore the partial derivatives of the function f(y, z) = (sin y)e^{z^3 - z - 1} + e^{yz}. This step-by-step solution covers key concepts in multivariable calculus such as the chain rule, product rule, and partial derivatives. Suitable for college-level students.

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