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Second partial derivatives Find the four second partial derivatives of the following functions. $$f(x, y)=x^{2} \sin y$$

Short Answer

Expert verified
Question: Compute the four second partial derivatives of the function $$f(x, y) = x^2 \sin y$$. Answer: The four second partial derivatives of the given function are: 1. $$f_{xx}(x, y) = 2 \sin y$$ 2. $$f_{yy}(x, y) = -x^2 \sin y$$ 3. $$f_{xy}(x, y) = 2x \cos y$$ 4. $$f_{yx}(x, y) = 2x \cos y$$

Step by step solution

01

Find the first partial derivatives

Firstly, we need to determine the first partial derivatives with respect to x and y: $$f_x(x, y) = \frac{\partial}{\partial x} (x^{2} \sin y) = 2x \sin y$$ $$f_y(x, y) = \frac{\partial}{\partial y} (x^{2} \sin y) = x^{2} \cos y$$
02

Find the second partial derivatives

Now, we will find the second partial derivatives using the first partial derivatives we found in step 1. 1. Second derivative with respect to x: $$f_{xx}(x, y) = \frac{\partial}{\partial x} (2x \sin y) = 2 \sin y$$ 2. Second derivative with respect to y: $$f_{yy}(x, y) = \frac{\partial}{\partial y} (x^{2} \cos y) = -x^{2} \sin y$$ 3. Second derivative with respect to x and y: $$f_{xy}(x, y) = \frac{\partial}{\partial y} (2x \sin y) = 2x \cos y$$ 4. Second derivative with respect to y and x: $$f_{yx}(x, y) = \frac{\partial}{\partial x} (x^{2} \cos y) = 2x\cos y$$ The four second partial derivatives are: 1. $$f_{xx}(x, y) = 2 \sin y$$ 2. $$f_{yy}(x, y) = -x^{2} \sin y$$ 3. $$f_{xy}(x, y) = 2x \cos y$$ 4. $$f_{yx}(x, y) = 2x \cos y$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives play a pivotal role in multivariable calculus, particularly when analyzing the behavior of functions with more than one independent variable. When working with a function like \(f(x, y) = x^{2} \text{sin} y\), determining the rate at which \(f\) changes with respect to each variable individually is crucial. This rate of change is captured by taking partial derivatives.

For instance, when we compute \(f_x(x, y)\), it represents the slope of the surface created by \(f\) along the direction of the \(x\)-axis, while keeping \(y\) constant. Similarly, the derivative \(f_y(x, y)\) represents the slope along the \(y\)-axis, with \(x\) held constant. Understanding these derivatives helps visualize the shape and gradient of the function's graph in the spatial plane.

Knowing the concept of the first partial derivatives is essential before one can delve into second partial derivatives, which provide information about the curvature and concavity of the surface derived from the function.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of several variables. The function \(f(x, y) = x^{2} \text{sin} y\) from our exercise is an example of a multivariable function since it depends on two independent variables, \(x\) and \(y\). In the realm of multivariable calculus, partial derivatives are just the first step towards understanding more complex behaviors of such functions.

Second partial derivatives, such as \(f_{xx}(x, y)\) and \(f_{yy}(x, y)\), further illuminate how the function's graph bends in different directions. If one imagines the surface that such a function represents, second partial derivatives can indicate whether the surface has a hill, valley, or saddle point nearby a specific point.

Additionally, analyzing second partial derivatives is fundamental when performing optimization tasks, for solving problems regarding local maxima and minima—the peaks and troughs of the function's graph.
Chain Rule in Calculus
The chain rule is a fundamental theorem in calculus that allows us to differentiate composite functions—functions made up by chaining together other functions. In the context of multivariable calculus, the chain rule becomes even more powerful as it lets us differentiate functions with respect to multiple variables that are themselves functions of other variables.

For instance, if one needs to find the rate of change of \(f(x(t), y(t))\) with respect to time \(t\), where \(x\) and \(y\) are functions of \(t\), we would use the multivariable chain rule. It combines the derivatives of \(f\) with respect to \(x\) and \(y\), and the derivatives of \(x\) and \(y\) with respect to \(t\), giving us a full picture of how \(f\) changes as \(t\) progresses.

This is essential when dealing with physical problems where variables depend on time or other parameters, allowing us to track how changes propagate through a system. Understanding the chain rule is vital for students to effectively tackle problems involving partial derivatives and optimize functions defined by multivariable systems.

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Most popular questions from this chapter

Surface area of a cone A cone with height \(h\) and radius \(r\) has a lateral surface area (the curved surface only, excluding the base) of \(S=\pi r \sqrt{r^{2}+h^{2}}\) a. Estimate the change in the surface area when \(r\) increases from \(r=2.50\) to \(r=2.55\) and \(h\) decreases from \(h=0.60\) to \(h=0.58\) b. When \(r=100\) and \(h=200,\) is the surface area more sensitive to a small change in \(r\) or a small change in \(h ?\) Explain.

Second Derivative Test Suppose the conditions of the Second Derivative Test are satisfied on an open disk containing the point \((a, b) .\) Use the test to prove that if \((a, b)\) is a critical point of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b) < 0 < f_{\text {vy }}(a, b)\) or \(f_{y y}(a, b) < 0 < f_{x x}(a, b),\) then \(f\) has a saddle point at \((a, b)\)

Travel cost The cost of a trip that is \(L\) miles long, driving a car that gets \(m\) miles per gallon, with gas costs of \(\$ p /\) gal is \(C=L p / m\) dollars. Suppose you plan a trip of \(L=1500 \mathrm{mi}\) in a car that gets \(m=32 \mathrm{mi} / \mathrm{gal},\) with gas costs of \(p=\$ 3.80 / \mathrm{gal}\) a. Explain how the cost function is derived. b. Compute the partial derivatives \(C_{L}, C_{m^{\prime}}\) and \(C_{p^{\prime}}\). Explain the meaning of the signs of the derivatives in the context of this problem. c. Estimate the change in the total cost of the trip if \(L\) changes from \(L=1500\) to \(L=1520, m\) changes from \(m=32\) to \(m=31,\) and \(p\) changes from \(p=\$ 3.80\) to \(p=\$ 3.85\) d. Is the total cost of the trip (with \(L=1500 \mathrm{mi}, m=32 \mathrm{mi} / \mathrm{gal}\). and \(p=\$ 3.80\) ) more sensitive to a \(1 \%\) change in \(L,\) in \(m,\) or in \(p\) (assuming the other two variables are fixed)? Explain.

Find the domain of the following functions. If possible, give a description of the domains (for example, all points outside a sphere of radius 1 centered at the origin). $$Q(x, y, z)=\frac{10}{1+x^{2}+y^{2}+4 z^{2}}$$

Challenge domains Find the domain of the following functions. Specify the domain mathematically, and then describe it in words or with a sketch. $$g(x, y, z)=\frac{10}{x^{2}-(y+z) x+y z}$$

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