Find both first partial derivatives. $$ h(x, y)=e^{-\left(x^{5}+y^{5}\right)} $$

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We are given

$$
h(x, y)=e^{-\left(x^{5}+y^{5}\right)}
$$

**Step 1: Find the partial derivative with respect to $$ x $$.**

Recall that when differentiating an exponential function, we apply the chain rule. For a function of the form $$ e^{g(x,y)} $$, the partial derivative with respect to $$ x $$ is

$$
\frac{\partial}{\partial x} e^{g(x,y)} = e^{g(x,y)} \cdot \frac{\partial}{\partial x} g(x,y)
$$

Here, the inner function is

$$
g(x,y) = -\left(x^5 + y^5\right)
$$

Differentiate $$ g(x,y) $$ with respect to $$ x $$:

$$
\frac{\partial}{\partial x}g(x,y) = -\frac{\partial}{\partial x}\left(x^5 + y^5\right) = -\left(5x^4 + 0\right) = -5x^4
$$

Thus, the partial derivative of $$ h $$ with respect to $$ x $$ is

$$
h_x(x, y)= e^{-\left(x^5+y^5\right)} \cdot (-5x^4) = -5x^4 e^{-\left(x^5+y^5\right)}
$$

**Step 2: Find the partial derivative with respect to $$ y $$.**

Similarly, apply the chain rule for the derivative with respect to $$ y $$:

$$
h_y(x, y) = e^{-\left(x^5+y^5\right)} \cdot \frac{\partial}{\partial y}\left[-\left(x^5+y^5\right)\right]
$$

Differentiate $$ g(x,y) $$ with respect to $$ y $$:

$$
\frac{\partial}{\partial y}g(x,y) = -\frac{\partial}{\partial y}\left(x^5+y^5\right) = -\left(0 + 5y^4\right) = -5y^4
$$

Thus, the partial derivative of $$ h $$ with respect to $$ y $$ is

$$
h_y(x, y)= e^{-\left(x^5+y^5\right)} \cdot (-5y^4) = -5y^4 e^{-\left(x^5+y^5\right)}
$$

**Final Answers:**

$$
h_x(x, y) = -5x^4 e^{-\left(x^5+y^5\right)}
$$

$$
h_y(x, y) = -5y^4 e^{-\left(x^5+y^5\right)}
$$
The first partial derivative with respect to $$ x $$ is $$ h_x(x, y) = -5x^4 e^{-\left(x^5+y^5\right)} $$, and the first partial derivative with respect to $$ y $$ is $$ h_y(x, y) = -5y^4 e^{-\left(x^5+y^5\right)} $$.

Find both first partial derivatives.

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