Problem 14 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$h(x, y)=\left(y^{2}+1\right) e^{x}$$

Step-by-Step Solution

Answer

Answer: The first partial derivative with respect to x is $$\frac{\partial h}{\partial x} = (y^2 + 1)e^x$$, and the first partial derivative with respect to y is $$\frac{\partial h}{\partial y} = 2y\cdot e^x$$.

Step 1: Differentiate with respect to x

To find the partial derivative with respect to x, treat y as a constant and differentiate the function with respect to x. $$\frac{\partial h}{\partial x} = \frac{\partial}{\partial x}((y^2+1)e^x)$$ Since y is a constant, we can apply the product rule here. $$\frac{\partial h}{\partial x} = (y^2 + 1)\cdot \frac{\partial}{\partial x}(e^x)$$ Now we can differentiate $$e^x$$ with respect to x. $$\frac{\partial h}{\partial x} = (y^2 + 1)\cdot e^x$$

Step 2: Differentiate with respect to y

To find the partial derivative with respect to y, we treat x as a constant and differentiate the function with respect to y. $$\frac{\partial h}{\partial y} = \frac{\partial}{\partial y}((y^2+1)e^x)$$ Since x is a constant, we can apply the product rule here. $$\frac{\partial h}{\partial y} = \frac{\partial}{\partial y}(y^2 + 1)\cdot e^x$$ Now we can differentiate $$y^2 + 1$$ with respect to y. $$\frac{\partial h}{\partial y} = 2y\cdot e^x$$

Step 3: Final answer

The first partial derivatives of the given function $$h(x, y) = (y^2+1)e^x$$ are: $$\frac{\partial h}{\partial x} = (y^2 + 1)e^x$$ and $$\frac{\partial h}{\partial y} = 2y\cdot e^x$$