We are given:

$z = e^{5x + y}$ $x = 6st$ $y = 7s + 6t$

We need to find $z_s$ (the partial derivative of $z$ with respect to $s$) and $z_t$ (the partial derivative of $z$ with respect to $t$).

Step 1: Compute $z_s$

By the chain rule, we have:

$z_s = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$

• First, calculate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$:

$\frac{\partial z}{\partial x} = 5e^{5x + y}, \quad \frac{\partial z}{\partial y} = e^{5x + y}$

• Next, compute $\frac{\partial x}{\partial s}$ and $\frac{\partial y}{\partial s}$:

$\frac{\partial x}{\partial s} = 6t, \quad \frac{\partial y}{\partial s} = 7$

Now, substitute everything into the equation for $z_s$:

$z_s = 5e^{5x + y} \cdot 6t + e^{5x + y} \cdot 7$

Factor out $e^{5x + y}$:

$z_s = e^{5x + y}(30t + 7)$

Step 2: Compute $z_t$

Similarly, by the chain rule:

$z_t = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}$

• We already know $\frac{\partial z}{\partial x} = 5e^{5x + y}$ and $\frac{\partial z}{\partial y} = e^{5x + y}$.

• Now compute $\frac{\partial x}{\partial t}$ and $\frac{\partial y}{\partial t}$:

$\frac{\partial x}{\partial t} = 6s, \quad \frac{\partial y}{\partial t} = 6$

Substitute everything into the equation for $z_t$:

$z_t = 5e^{5x + y} \cdot 6s + e^{5x + y} \cdot 6$

Factor out $e^{5x + y}$:

$z_t = e^{5x + y}(30s + 6)$

Final Answers:

$z_s = e^{5x + y}(30t + 7)$ $z_t = e^{5x + y}(30s + 6)$

Would you like a more detailed explanation or have any questions?

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Relative Questions:

1. How does the chain rule apply when differentiating functions of multiple variables?
2. Can you provide a general form for partial derivatives using the chain rule?
3. What is the physical interpretation of partial derivatives in a multivariable context?
4. How do exponential functions behave when differentiated?
5. How do product and chain rules combine in more complex derivatives?

Tip: When differentiating, always look for opportunities to factor common terms for simpler results.