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Second partial derivatives Find the four second partial derivatives of the following functions. $$f(x, y)=y^{3} \sin 4 x$$

Short Answer

Expert verified
Question: Find the four second partial derivatives of the function $$f(x, y) = y^3 \sin 4x$$. Answer: The four second partial derivatives of the function $$f(x, y) = y^3 \sin 4x$$ are: 1. $$f_{xx}(x, y) = -16y^3 \sin 4x$$ 2. $$f_{yy}(x, y) = 6y \sin 4x$$ 3. $$f_{xy}(x, y) = 12y^2 \cos 4x$$ 4. $$f_{yx}(x, y) = 12y^2 \cos 4x$$

Step by step solution

01

Find the first partial derivatives with respect to x and y

To find the first partial derivatives, we will differentiate the function f(x, y) with respect to x and y: $$f_x(x, y) = \frac{\partial}{\partial x} (y^3 \sin 4x)$$ $$f_y(x, y) = \frac{\partial}{\partial y} (y^3 \sin 4x)$$ After differentiating, we get: $$f_x(x, y) = 4y^3 \cos 4x$$ $$f_y(x, y) = 3y^2 \sin 4x$$
02

Find the second partial derivatives with respect to x twice, $$f_{xx}(x, y)$$

Differentiate the first partial derivative with respect to x again: $$f_{xx}(x, y) = \frac{\partial^2}{\partial x^2} (y^3 \sin 4x) = \frac{\partial}{\partial x} (4y^3 \cos 4x)$$ After differentiating, we get: $$f_{xx}(x, y) = -16y^3 \sin 4x$$
03

Find the second partial derivatives with respect to y twice, $$f_{yy}(x, y)$$

Differentiate the first partial derivative with respect to y again: $$f_{yy}(x, y) = \frac{\partial^2}{\partial y^2} (y^3 \sin 4x) = \frac{\partial}{\partial y} (3y^2 \sin 4x)$$ After differentiating, we get: $$f_{yy}(x, y) = 6y \sin 4x$$
04

Find the second partial derivatives with respect to x and y, $$f_{xy}(x, y)$$

Differentiate the first partial derivative with respect to x first and then y: $$f_{xy}(x, y) = \frac{\partial^2}{\partial x \partial y} (y^3 \sin 4x) = \frac{\partial}{\partial y} (4y^3 \cos 4x)$$ After differentiating, we get: $$f_{xy}(x, y) = 12y^2 \cos 4x$$
05

Find the second partial derivatives with respect to y and x, $$f_{yx}(x, y)$$

Differentiate the first partial derivative with respect to y first and then x: $$f_{yx}(x, y) = \frac{\partial^2}{\partial y \partial x} (y^3 \sin 4x) = \frac{\partial}{\partial x} (3y^2 \sin 4x)$$ After differentiating, we get: $$f_{yx}(x, y) = 12y^2 \cos 4x$$ Thus, the four second partial derivatives are: $$f_{xx}(x, y) = -16y^3 \sin 4x$$ $$f_{yy}(x, y) = 6y \sin 4x$$ $$f_{xy}(x, y) = 12y^2 \cos 4x$$ $$f_{yx}(x, y) = 12y^2 \cos 4x$$

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Key Concepts

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

Partial Differentiation
Partial differentiation is a fundamental technique used in multivariable calculus to determine the rate at which a function changes with respect to one of the variables, while keeping the other variables constant. It is represented by the symbol \( \frac{\partial}{\partial x} \) for differentiating with respect to the variable \( x \) and \( \frac{\partial}{\partial y} \) for differentiating with respect to \( y \).

Considering the example \( f(x, y) = y^{3} \sin 4x \), the first partial derivatives, \( f_x \) and \( f_y \) reflect changes in \( f \) as \( x \) or \( y \) vary independently. The derivatives \( f_x(x, y) = 4y^3 \cos 4x \) and \( f_y(x, y) = 3y^2 \sin 4x \) tell us how the function's slope behaves in the direction of each axis when the other variable is held constant. Understanding partial differentiation is key to analyzing the behavior of multivariable functions in different directions.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. Unlike single-variable calculus, where only one independent variable changes, multivariable calculus deals with functions like \( f(x, y) \) which depends on two or more variables.

In a function such as \( f(x, y) \) mentioned in the exercise, we investigate how the function behaves as each variable changes, and how it is affected by the simultaneous change of both variables. This understanding is crucial for fields like physics, engineering, and economics, where several factors influence a single outcome. The four second partial derivatives calculated in the exercise exhibit how the rates of change interact with each other in different combinations of \( x \) and \( y \).
Chain Rule in Partial Differentiation
The chain rule in partial differentiation is a powerful tool for finding the derivative of a function with respect to one variable when that function is composed of other functions which themselves are dependent on that variable. It comes into play especially when we deal with functions within functions, or 'nested' functions.

Analogous to the basic chain rule from single-variable calculus, the multivariable chain rule allows us to dissect complex relationships by calculating the derivatives of the outer and inner functions and then appropriately multiplying them. This rule greatly simplifies the process of finding derivatives in complex situations, and applying it correctly is vital for the accurate analysis of multivariable functions.
Higher-Order Derivatives
Higher-order derivatives refer to taking the derivative of a function multiple times. When working with multivariable functions like \( f(x, y) \) seen in our exercise, we can take the second, third, and higher-order derivatives with respect to each variable.

The second partial derivatives \( f_{xx} \) and \( f_{yy} \) give us the curve's concavity in their respective variable directions. Cross partial derivatives, such as \( f_{xy} \) and \( f_{yx} \) indicate how the slope in one direction changes as we move along another direction, and surprisingly in most cases, \( f_{xy} \) equals \( f_{yx} \) due to Clairaut's Theorem on equality of mixed partials—This is seen in our exercise as well. Understanding higher-order derivatives is vital for approximating functions and studying the geometry of surfaces in a multivariable context.

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Most popular questions from this chapter

Suppose the elevation of Earth's surface over a 16 -mi by 16 -mi region is approximated by the function \(z=10 e^{-\left(x^{2}+y^{2}\right)}+5 e^{-\left((x+5)^{2}+(y-3)^{2}\right) / 10}+4 e^{-2\left((x-4)^{2}+(y+1)^{2}\right)}\) a. Graph the height function using the window \([-8,8] \times[-8,8] \times[0,15]\) b. Approximate the points \((x, y)\) where the peaks in the landscape appear. c. What are the approximate elevations of the peaks?

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Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

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