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Chapter 13: Problem 19

Find both first partial derivatives. $$ h(x, y)=e^{-\left(x^{2}+y^{2}\right)} $$

Short Answer

Expert verified
The first partial derivatives of the function are \( h_x = -2x * e^{-x^2 - y^2} \) and \( h_y = -2y * e^{-x^2 - y^2} \)

Step by step solution

01

Find the Partial Derivative with respect to x

The partial derivative of the function \( h(x, y) \) with respect to x, denoted as \( h_x \), represents the rate at which the function is changing with respect to x, assuming y is held constant. To compute \( h_x \), apply the chain rule: the derivative of the outer function times the derivative of the inner function. \n The outer function is \( e^{u} \), its derivative is also \( e^{u} \). Here, \( u = -x^2 - y^2 \). So \( du = -2x \) dx (holding y constant). \n Therefore, \n \( h_x = e^{u} * du = -2x * e^{-x^2 - y^2} \)
02

Find the Partial Derivative with respect to y

Proceed similarly for the partial derivative with respect to y. The outer and inner functions remain the same, but now \( du = -2y \) dy (holding x constant). So, \n \( h_y = e^{u} * du = -2y * e^{-x^2 - y^2} \)

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Key Concepts

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

Chain Rule in Partial Derivatives
The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. In the context of partial derivatives, the chain rule helps us find how a function changes with respect to one variable while holding others constant.

For a function like \( h(x, y) = e^{-x^2 - y^2} \), to find the partial derivative with respect to a single variable, such as \( x \) or \( y \), we treat the other variables as constants. The chain rule tells us that to differentiate the entire function, we:
  • Identify the inside function, often known as the inner function, in this case, \( -x^2 - y^2 \).

  • Differentiated with respect to the chosen variable (either \( x \) or \( y \)).

  • Multiply the derivative of the outer function by the derivative of the inner function.
This method simplifies the process of finding partial derivatives in functions involving multiple variables.
Function Differentiation
Function differentiation is the process of finding the derivative or rate of change of a function. In multi-variable calculus, this means computing partial derivatives. Differentiating a function tells us how it changes with respect to an infinitesimally small change in one of its variables, while keeping the others fixed.

To compute partial derivatives for a function like \( h(x, y) = e^{-x^2 - y^2} \):
  • First, determine which variable you wish to differentiate with respect to 鈥� \( x \) for \( h_x \) or \( y \) for \( h_y \).

  • Calculate the derivative of the inner function, which forms part of a composite function.

  • Apply the chain rule to complete the differentiation process.
By performing these steps, you obtain the partial derivative that indicates how the function \( h(x, y) \) changes as you move along the direction defined by a specific variable.
Exponential Function Differentiation
Exponential functions, with bases as numbers such as \( e \), pose unique advantages in differentiation. One key property is that the derivative of an exponential function \( e^u \), where \( u \) is a function of \( x \) or \( y \), is simply \( e^u \) itself multiplied by the derivative of \( u \).

For our function \( h(x, y) = e^{-x^2 - y^2} \):
  • The base of the exponential function is the natural constant \( e \). The expression \( -x^2 - y^2 \) acts as the exponent.

  • The derivative involves multiplying \( e^{-x^2 - y^2} \) by the derivative of \( -x^2 - y^2 \) with respect to the variable of interest.

  • This results in \( -2x \cdot e^{-x^2 - y^2} \) when differentiating with respect to \( x \), or \( -2y \cdot e^{-x^2 - y^2} \) when differentiating with respect to \( y \).
This property greatly simplifies the differentiation of complex exponential functions, making such processes straightforward once the inner function is identified.

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Most popular questions from this chapter

Use Lagrange multipliers to find the indicated extrema, assuming that \(x\) and \(y\) are positive. Minimize \(f(x, y)=x^{2}-y^{2}\) Constraint: \(x-2 y+6=0\)

Find symmetric equations of the tangent line to the curve of intersection of the surfaces at the given point, and (b) find the cosine of the angle between the gradient vectors at this point. State whether or not the surfaces are orthogonal at the point of intersection.\(z=x^{2}+y^{2}, \quad x+y+6 z=33, \quad(1,2,5)\)

Find the absolute extrema of the function over the region \(R .\) (In each case, \(R\) contains the boundaries.) Use a computer algebra system to confirm your results. \(f(x, y)=(2 x-y)^{2}\) \(R\) : The triangular region in the \(x y\) -plane with vertices \((2,0)\), \((0,1)\), and \((1,2)\)

Use Lagrange multipliers to find the dimensions of a right circular cylinder with volume \(V_{0}\) cubic units and minimum surface area.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The graph of \(f(x, y)=x^{2}-y^{2}\) is a hyperbolic paraboloid.
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