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Find the four second partial derivatives of the following functions. $$h(x, y)=x^{3}+x y^{2}+1$$

Short Answer

Expert verified
Using the function \(h(x, y) = x^3 + xy^2 + 1\), find the four second partial derivatives. Solution: The four second partial derivatives are: 1) \(\frac{\partial^2 h}{\partial x^2} = 6x\) 2) \(\frac{\partial^2 h}{\partial y^2} = 2x\) 3) \(\frac{\partial^2 h}{\partial x \partial y} = 2y\) 4) \(\frac{\partial^2 h}{\partial y \partial x} = 2y\)

Step by step solution

01

Find the first partial derivative with respect to x

The first partial derivative of \(h(x, y)\) with respect to \(x\) is: $$ \frac{\partial h}{\partial x} = \frac{\partial}{\partial x}(x^3 + xy^2 + 1) $$ Apply the derivatives: $$ \frac{\partial h}{\partial x} = 3x^2+y^2 $$
02

Find the first partial derivative with respect to y

The first partial derivative of \(h(x, y)\) with respect to \(y\) is: $$ \frac{\partial h}{\partial y} = \frac{\partial}{\partial y}(x^3 + xy^2 + 1) $$ Apply the derivatives: $$ \frac{\partial h}{\partial y} = 2xy $$
03

Find the second partial derivative \(\frac{\partial^2 h}{\partial x^2}\)

Now we will find the second partial derivative of \(h(x, y)\) with respect to \(x\): $$ \frac{\partial^2 h}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial h}{\partial x}\right) = \frac{\partial}{\partial x}(3x^2+y^2) $$ Apply the derivatives: $$ \frac{\partial^2 h}{\partial x^2} = 6x $$
04

Find the second partial derivative \(\frac{\partial^2 h}{\partial y^2}\)

Now we will find the second partial derivative of \(h(x, y)\) with respect to \(y\): $$ \frac{\partial^2 h}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial h}{\partial y}\right) = \frac{\partial}{\partial y}(2xy) $$ Apply the derivatives: $$ \frac{\partial^2 h}{\partial y^2} = 2x $$
05

Find the mixed partial derivative \(\frac{\partial^2 h}{\partial x \partial y}\)

Now we will find the mixed partial derivative of \(h(x, y)\) with respect to \(x\) and \(y\): $$ \frac{\partial^2 h}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial h}{\partial y}\right) = \frac{\partial}{\partial x}(2xy) $$ Apply the derivatives: $$ \frac{\partial^2 h}{\partial x \partial y} = 2y $$
06

Find the mixed partial derivative \(\frac{\partial^2 h}{\partial y \partial x}\)

Now we will find the mixed partial derivative of \(h(x, y)\) with respect to \(y\) and \(x\): $$ \frac{\partial^2 h}{\partial y \partial x} = \frac{\partial}{\partial y}\left(\frac{\partial h}{\partial x}\right) = \frac{\partial}{\partial y}(3x^2+y^2) $$ Apply the derivatives: $$ \frac{\partial^2 h}{\partial y \partial x} = 2y $$ In conclusion, the four second partial derivatives are: 1) \(\frac{\partial^2 h}{\partial x^2} = 6x\) 2) \(\frac{\partial^2 h}{\partial y^2} = 2x\) 3) \(\frac{\partial^2 h}{\partial x \partial y} = 2y\) 4) \(\frac{\partial^2 h}{\partial y \partial x} = 2y\)

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Key Concepts

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

second partial derivatives
A second partial derivative is essentially the derivative of a derivative. This means you take the derivative of your function with respect to one variable and then derive that result again with respect to the same variable. For the given function \( h(x, y) = x^3 + xy^2 + 1 \), finding the second partial derivatives involves computing the derivatives twice.
For the derivative \( \frac{\partial^2 h}{\partial x^2} \), you first compute \( \frac{\partial h}{\partial x} = 3x^2 + y^2 \) and then differentiate again with respect to \( x \), resulting in:
  • \( \frac{\partial^2 h}{\partial x^2} = 6x \)
This tells us how the rate of change of the function, with respect to \( x \), changes as \( x \) itself changes more. Similarly, for \( \frac{\partial^2 h}{\partial y^2} \), you first compute \( \frac{\partial h}{\partial y} = 2xy \) and repeat the process with respect to \( y \), leading to:
  • \( \frac{\partial^2 h}{\partial y^2} = 2x \)
Second partial derivatives are especially useful in determining the concavity or curvature of a function's graph.
mixed partial derivatives
Mixed partial derivatives involve taking the partial derivative of a function with respect to two different variables. They are crucial in multivariable calculus because they show how the function changes as multiple variables change together.For our function, \( h(x, y) = x^3 + xy^2 + 1 \), there are two mixed partial derivatives: \( \frac{\partial^2 h}{\partial x \partial y} \) and \( \frac{\partial^2 h}{\partial y \partial x} \). These measure changes through a sequence of differentiation - first in one direction, then in another.First, let's look at \( \frac{\partial^2 h}{\partial x \partial y} \):
  • First, find \( \frac{\partial h}{\partial y} = 2xy \) and then differentiate with respect to \( x \) to get \( \frac{\partial^2 h}{\partial x \partial y} = 2y \).
Similarly, let's consider \( \frac{\partial^2 h}{\partial y \partial x} \):
  • First, find \( \frac{\partial h}{\partial x} = 3x^2 + y^2 \) and then differentiate with respect to \( y \) to again obtain \( \frac{\partial^2 h}{\partial y \partial x} = 2y \).
Notice that these mixed derivatives are equal, which is a consistent result according to Clairaut's theorem on the symmetry of second partial derivatives.
multivariable calculus
Multivariable calculus extends the principles of calculus from single-variable functions to functions of several variables. It's foundational for fields that require modeling of systems dependent on several factors.Key concepts include:
  • Partial Derivatives: These represent how a function changes as only one variable changes, keeping others constant. For instance, \( \frac{\partial h}{\partial x} \) conveys how \( h \) alters as \( x \) changes, treating \( y \) as constant.

  • Second Partial Derivatives: Represent how the rate of change itself changes, which helps determine curvature properties of surfaces modeled by the functions.

  • Mixed Partial Derivatives: These explore changes in different directions, offering insight into interactions between variables.
A thorough understanding of these concepts allows one to analyze and solve complex problems spanning physics, engineering, economics, and beyond. In our example with \( h(x, y) \), we've seen how these derivatives are calculated and what they signify, demonstrating the fundamental role of multivariable calculus.

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