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

To find the first partial derivatives of the function \( h(x, y) = e^{-(x^5 + y^5)} \), we can apply the chain rule. First, for the partial derivative with respect to \( x \): \[ \frac{\partial h}{\partial x} = \frac{\partial}{\partial x} \left(e^{-(x^5 + y^5)}\right) \] Using the chain rule: \[ = e^{-(x^5 + y^5)} \cdot \frac{\partial}{\partial x} (-(x^5 + y^5)) = e^{-(x^5 + y^5)} \cdot (-5x^4) \] Thus, we have: \[ \frac{\partial h}{\partial x} = -5x^4 e^{-(x^5 + y^5)} \] Next, for the partial derivative with respect to \( y \): \[ \frac{\partial h}{\partial y} = \frac{\partial}{\partial y} \left(e^{-(x^5 + y^5)}\right) \] Similarly: \[ = e^{-(x^5 + y^5)} \cdot \frac{\partial}{\partial y} (-(x^5 + y^5)) = e^{-(x^5 + y^5)} \cdot (-5y^4) \] Thus: \[ \frac{\partial h}{\partial y} = -5y^4 e^{-(x^5 + y^5)} \] In summary, the first partial derivatives are: \[ \frac{\partial h}{\partial x} = -5x^4 e^{-(x^5 + y^5)} \] \[ \frac{\partial h}{\partial y} = -5y^4 e^{-(x^5 + y^5)} \]