Part 8 of 8 Points: 0 of 1 Find the following derivatives. $$ z_{s} $$ and $$ z_{4} $$, where $$ z=e^{3 x+5 y}, x=8 s t $$, and $$ y=6 s+5 t $$ $$ \frac{\partial z}{\partial y}=5 e^{3 x+5 y} $$ (Type an expression using x and y as the variables.) $$ \frac{\partial y}{\partial s}=\overline{6} $$ (Type an expression using $$ s $$ and $$ t $$ as the variables.) $$ \frac{\partial x}{\partial t}=\frac{}{8 s} $$ (Type an expression using $$ s $$ and $$ t $$ as the variables.) $$ \frac{\partial y}{\partial t}=5 $$ (Type an expression using $$ s $$ and $$ t $$ as the variables.) $$ z_{s}=(24 t+30) e^{24 s t+30 s+25 t} $$ (Type an expression using $$ s $$ and $$ t $$ as the variables.) $$ z_{1}=\square $$ (Type an expression using $$ s $$ and $$ t $$ as the variables.)

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UpStudy AI · Accepted Answer

Given
$$
z=e^{3x+5y},\quad x=8st,\quad y=6s+5t,
$$
we first note that
$$
z=e^{u}\quad\text{with}\quad u=3x+5y.
$$

**Step 1.** Compute $$ u $$ in terms of $$ s $$ and $$ t $$:
$$
u=3(8st)+5(6s+5t)=24st+30s+25t.
$$

**Step 2.** Find $$ z_s $$ using the chain rule:
$$
z_s=\frac{\partial z}{\partial s}=e^{u}\frac{\partial u}{\partial s}.
$$
Differentiate $$ u=24st+30s+25t $$ with respect to $$ s $$:
$$
\frac{\partial u}{\partial s}=24t+30.
$$
Thus,
$$
z_s=(24t+30)e^{24st+30s+25t}.
$$

**Step 3.** Compute $$ z_t $$ (interpreting the notation $$ z_1 $$ as the derivative with respect to $$ t $$):
$$
z_t=\frac{\partial z}{\partial t}=e^{u}\frac{\partial u}{\partial t}.
$$
Differentiate $$ u=24st+30s+25t $$ with respect to $$ t $$:
$$
\frac{\partial u}{\partial t}=24s+25.
$$
Thus,
$$
z_t=(24s+25)e^{24st+30s+25t}.
$$

The final answers are:
$$
z_s=(24t+30)e^{24st+30s+25t}\quad\text{and}\quad z_t=(24s+25)e^{24st+30s+25t}.
$$
$$ z_s = (24t + 30)e^{24st + 30s + 25t} $$ and $$ z_t = (24s + 25)e^{24st + 30s + 25t} $$.

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