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Chapter 13: Problem 14

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

Short Answer

Expert verified
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 by step solution

01

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. $$\frac{\partial h}{\partial x} = (y^2 + 1)\cdot e^x$$
02

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. $$\frac{\partial h}{\partial y} = 2y\cdot e^x$$
03

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 $$\frac{\partial h}{\partial y} = 2y\cdot e^x$$

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Most popular questions from this chapter

Suppose \(P\) is a point in the plane \(a x+b y+c z=d .\) Then the least distance from any point \(Q\) to the plane equals the length of the orthogonal projections of \(\overrightarrow{P Q}\) onto the normal vector \(\mathbf{n}=\langle a, b . c\rangle\) a. Use this information to show that the least distance from \(Q\) to the plane is \(\frac{|\overrightarrow{P Q} \cdot \mathbf{n}|}{|\mathbf{n}|}\) b. Find the least distance from the point (1,2,-4) to the plane \(2 x-y+3 z=1\)

Consider the ellipse \(x^{2}+4 y^{2}=1\) in the \(x y\) -plane. a. If this ellipse is revolved about the \(x\) -axis, what is the equation of the resulting ellipsoid? b. If this ellipse is revolved about the \(y\) -axis, what is the equation of the resulting ellipsoid?

Let \(w=f(x, y, z)=2 x+3 y+4 z\), which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\), \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\).

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