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Chapter 12: Problem 48

Find the first partial derivatives of the following functions. $$Q(x, y, z)=\tan x y z$$

Short Answer

Expert verified
The first partial derivatives are: - With respect to x: $$\frac{\partial Q}{\partial x} = (\sec^2{x y z})(yz)$$ - With respect to y: $$\frac{\partial Q}{\partial y} = (\sec^2{x y z})(xz)$$ - With respect to z: $$\frac{\partial Q}{\partial z} = (\sec^2{x y z})(xy)$$

Step by step solution

01

Find the partial derivative with respect to x

To find the partial derivative of the function Q with respect to x, we can treat y and z as constants: $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(\tan{x y z})$$ Now, we will use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function: $$\frac{\partial Q}{\partial x} = (\sec^2{x y z})(\frac{\partial}{\partial x}(x y z))$$ With y and z treated as constants, the partial derivative of the inner function x y z with respect to x is just yz: $$\frac{\partial Q}{\partial x} = (\sec^2{x y z})(yz)$$
02

Find the partial derivative with respect to y

Similarly, to find the partial derivative of the function Q with respect to y, we can treat x and z as constants: $$\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(\tan{x y z})$$ Using the chain rule again: $$\frac{\partial Q}{\partial y} = (\sec^2{x y z})(\frac{\partial}{\partial y}(x y z))$$ With x and z treated as constants, the partial derivative of the inner function x y z with respect to y is just xz: $$\frac{\partial Q}{\partial y} = (\sec^2{x y z})(xz)$$
03

Find the partial derivative with respect to z

Finally, to find the partial derivative of the function Q with respect to z, we can treat x and y as constants: $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(\tan{x y z})$$ Using the chain rule one more time: $$\frac{\partial Q}{\partial z} = (\sec^2{x y z})(\frac{\partial}{\partial z}(x y z))$$ With x and y treated as constants, the partial derivative of the inner function x y z with respect to z is just xy: $$\frac{\partial Q}{\partial z} = (\sec^2{x y z})(xy)$$ In summary, the first partial derivatives of the given function Q are: $$\frac{\partial Q}{\partial x} = (\sec^2{x y z})(yz)$$ $$\frac{\partial Q}{\partial y} = (\sec^2{x y z})(xz)$$ $$\frac{\partial Q}{\partial z} = (\sec^2{x y z})(xy)$$

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Most popular questions from this chapter

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{a x} \cos a y, \text { for any real number } a$$

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May 1985, and Calculus and Analytical Geometry, 2nd ed., Philip Gillett, 1984)

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?

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) In two dimensions, the motion of an ideal fluid (an incompressible and irrotational fluid) is governed by a velocity potential \(\varphi .\) The velocity components of the fluid, \(u\) in the \(x\) -direction and \(v\) in the \(y\) -direction, are given by \(\langle u, v\rangle=\nabla \varphi .\) Find the velocity components associated with the velocity potential \(\varphi(x, y)=\sin \pi x \sin 2 \pi y\).

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