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Find the second partial derivatives of \(f\). $$ f(x, y)=x^{3} y^{2}-3 x y^{3}+x^{4}-3 y+2 $$

Short Answer

Expert verified
The second partial derivatives are: \(f_{xx} = 6xy^2 + 12x^2\), \(f_{yy} = 2x^3 - 18xy\), \(f_{xy} = f_{yx} = 6x^2 y - 9y^2\).

Step by step solution

01

Find the first partial derivative with respect to x

To find the partial derivative of \(f(x, y)\) with respect to \(x\), treat \(y\) as a constant. Differentiate term by term:- \(\frac{\partial}{\partial x}(x^3 y^2) = 3x^2 y^2\)- \(\frac{\partial}{\partial x}(-3xy^3) = -3y^3\)- \(\frac{\partial}{\partial x}(x^4) = 4x^3\)- \(\frac{\partial}{\partial x}(-3y) = 0\) - \(\frac{\partial}{\partial x}(2) = 0\) So, \(f_x(x,y) = 3x^2 y^2 - 3y^3 + 4x^3\).
02

Find the first partial derivative with respect to y

To find the partial derivative of \(f(x, y)\) with respect to \(y\), treat \(x\) as a constant. Differentiate term by term:- \(\frac{\partial}{\partial y}(x^3 y^2) = 2x^3 y\)- \(\frac{\partial}{\partial y}(-3xy^3) = -9xy^2\)- \(\frac{\partial}{\partial y}(x^4) = 0\)- \(\frac{\partial}{\partial y}(-3y) = -3\)- \(\frac{\partial}{\partial y}(2) = 0\)So, \(f_y(x,y) = 2x^3 y - 9xy^2 - 3\).
03

Find the second partial derivative with respect to x twice

Differentiate \(f_x(x, y) = 3x^2 y^2 - 3y^3 + 4x^3\) with respect to \(x\) again:- \(\frac{\partial}{\partial x}(3x^2 y^2) = 6xy^2\)- \(\frac{\partial}{\partial x}(-3y^3) = 0\)- \(\frac{\partial}{\partial x}(4x^3) = 12x^2\)So, \(f_{xx}(x, y) = 6xy^2 + 12x^2\).
04

Find the second partial derivative with respect to y twice

Differentiate \(f_y(x, y) = 2x^3 y - 9xy^2 - 3\) with respect to \(y\) again:- \(\frac{\partial}{\partial y}(2x^3 y) = 2x^3\)- \(\frac{\partial}{\partial y}(-9xy^2) = -18xy\)- \(\frac{\partial}{\partial y}(-3) = 0\)So, \(f_{yy}(x, y) = 2x^3 - 18xy\).
05

Find the mixed partial derivative with respect to x first, then y

Differentiate \(f_x(x, y) = 3x^2 y^2 - 3y^3 + 4x^3\) with respect to \(y\):- \(\frac{\partial}{\partial y}(3x^2 y^2) = 6x^2 y\)- \(\frac{\partial}{\partial y}(-3y^3) = -9y^2\)- \(\frac{\partial}{\partial y}(4x^3) = 0\)So, \(f_{xy}(x, y) = 6x^2 y - 9y^2\).
06

Confirm the mixed partial derivative with respect to y first, then x

Differentiate \(f_y(x, y) = 2x^3 y - 9xy^2 - 3\) with respect to \(x\):- \(\frac{\partial}{\partial x}(2x^3 y) = 6x^2 y\)- \(\frac{\partial}{\partial x}(-9xy^2) = -9y^2\)- \(\frac{\partial}{\partial x}(-3) = 0\)So, \(f_{yx}(x, y) = 6x^2 y - 9y^2\), confirming that \(f_{xy} = f_{yx}\).

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Key Concepts

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

Second Partial Derivative
The second partial derivative is a concept from calculus that extends the idea of derivatives from single-variable functions to those of multiple variables. When working with functions of two variables like \(f(x, y)\), we can take derivatives with respect to \(x\) or \(y\).
Finding the second partial derivative involves differentiating a function twice with respect to the same variable. For example, if you first find the partial derivative of a function with respect to \(x\), you would then differentiate that result again with respect to \(x\) to obtain the second partial derivative.
In mathematical notation, this is denoted as \(f_{xx}(x, y)\). Similarly, if you first differentiate with respect to \(y\) and then again with respect to \(y\), the notation would be \(f_{yy}(x, y)\).
  • \(f_{xx}:\) Second derivative with respect to \(x\).
  • \(f_{yy}:\) Second derivative with respect to \(y\).
Calculating these second partial derivatives gives us information about the curvature of the function's graph in the direction of each variable. This can help in analyzing important features such as concavity and points of inflection.
Mixed Derivatives
Mixed derivatives involve taking the partial derivative of a function with respect to two different variables. For instance, starting with a function \(f(x, y)\), you can first differentiate with respect to \(x\), obtaining \(f_x(x, y)\), and then take the derivative of this result with respect to \(y\), leading to \(f_{xy}(x, y)\).
The opposite order, differentiating first with respect to \(y\) and then \(x\), is \(f_{yx}(x, y)\). A crucial theorem in multivariable calculus, known as Clairaut's Theorem or Schwarz's Theorem, states that for functions that are continuous and have continuous second partial derivatives, the order of differentiation doesn't matter. This means \(f_{xy} = f_{yx}\).
  • \(f_{xy}:\) First with \(x\), then with \(y\).
  • \(f_{yx}:\) First with \(y\), then with \(x\).
Understanding mixed derivatives gives insight into how two variables may interact within a function. This can be critical for finding concepts like the saddle points and understanding the behavior of the function's graph.
Calculus of Functions of Several Variables
Calculus of functions of several variables is an extension of single-variable calculus focusing on functions that have more than one input, such as \(f(x, y)\) or \(g(x, y, z)\). This field allows us to study surfaces and the spaces they occupy.
Key techniques include partial derivatives, which measure how functions change as one particular variable changes while others are held constant. Second-order and mixed partial derivatives further extend this analysis by examining curvature and variable interaction. Understanding these concepts requires a familiarity with the basic rules of differentiation, including the product and chain rules, but applied within a multivariable context.
  • Partial derivatives analyze the effect of changing one input while fixing others.
  • Second partial derivatives deliver second-order change insights, indicating curvature.
  • Mixed derivatives reveal interaction effects between variables.
The calculus of functions of several variables has wide applications in physics, engineering, economics, and more, making it a critical area of study.

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