Second Partial Derivative 99E8B7

1. **State the problem:** We need to find the second partial derivative of $$z$$ with respect to $$x$$, where $$z = \frac{\sinh(x^2 y) e^{-y^2}}{2}$$. 2. **Recall the formula and rules:** The second partial derivative $$\frac{\partial^2 z}{\partial x^2}$$ means we first find $$\frac{\partial z}{\partial x}$$ and then differentiate that result again with respect to $$x$$. 3. **Find the first partial derivative $$\frac{\partial z}{\partial x}$$:** Since $$z = \frac{1}{2} \sinh(x^2 y) e^{-y^2}$$, treat $$y$$ as a constant when differentiating with respect to $$x$$. Use the chain rule: $$\frac{\partial z}{\partial x} = \frac{1}{2} e^{-y^2} \cdot \frac{d}{dx} \sinh(x^2 y)$$ Recall $$\frac{d}{dx} \sinh(u) = \cosh(u) \cdot \frac{du}{dx}$$. Here, $$u = x^2 y$$, so $$\frac{du}{dx} = 2xy$$. Therefore, $$\frac{\partial z}{\partial x} = \frac{1}{2} e^{-y^2} \cdot \cosh(x^2 y) \cdot 2 x y = x y e^{-y^2} \cosh(x^2 y)$$. 4. **Find the second partial derivative $$\frac{\partial^2 z}{\partial x^2}$$:** Differentiate $$\frac{\partial z}{\partial x} = x y e^{-y^2} \cosh(x^2 y)$$ with respect to $$x$$. Since $$y$$ and $$e^{-y^2}$$ are constants with respect to $$x$$, apply the product rule: $$\frac{\partial^2 z}{\partial x^2} = y e^{-y^2} \cdot \frac{d}{dx} \left( x \cosh(x^2 y) \right)$$. Use product rule on $$x \cosh(x^2 y)$$: $$\frac{d}{dx} (x \cosh(x^2 y)) = \cosh(x^2 y) + x \cdot \frac{d}{dx} \cosh(x^2 y)$$. Recall $$\frac{d}{dx} \cosh(u) = \sinh(u) \cdot \frac{du}{dx}$$, with $$u = x^2 y$$ and $$\frac{du}{dx} = 2 x y$$. So, $$\frac{d}{dx} \cosh(x^2 y) = \sinh(x^2 y) \cdot 2 x y$$. Therefore, $$\frac{d}{dx} (x \cosh(x^2 y)) = \cosh(x^2 y) + x \cdot (\sinh(x^2 y) \cdot 2 x y) = \cosh(x^2 y) + 2 x^2 y \sinh(x^2 y)$$. 5. **Combine all parts:** $$\frac{\partial^2 z}{\partial x^2} = y e^{-y^2} \left( \cosh(x^2 y) + 2 x^2 y \sinh(x^2 y) \right)$$. 6. **Final answer:** $$\boxed{\frac{\partial^2 z}{\partial x^2} = y e^{-y^2} \left( \cosh(x^2 y) + 2 x^2 y \sinh(x^2 y) \right)}$$