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Chapter 11: Problem 18

\(7-30\) Find the first partial derivatives of the function. $$f(x, y)=x^{y}$$

Short Answer

Expert verified
The first partial derivatives are \( \frac{\partial f}{\partial x} = y \cdot x^{y-1} \) and \( \frac{\partial f}{\partial y} = x^y \cdot \ln(x) \).

Step by step solution

01

Understanding the Function

We are given the function \( f(x, y) = x^y \). This function depends on two variables, \( x \) and \( y \), so it has two first partial derivatives: one with respect to \( x \) and one with respect to \( y \). Partial derivatives are the derivatives of a function of multiple variables taken with respect to one variable, keeping the others constant.
02

Partial Derivative with Respect to x

To find the partial derivative of \( f(x, y) = x^y \) with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), we treat \( y \) as a constant. The expression for the derivative of \( x^y \) with respect to \( x \) is \( y \cdot x^{y-1} \). Hence, the partial derivative is: \[ \frac{\partial f}{\partial x} = y \cdot x^{y-1} \]
03

Partial Derivative with Respect to y

To find the partial derivative of \( f(x, y) = x^y \) with respect to \( y \), denoted \( \frac{\partial f}{\partial y} \), we treat \( x \) as a constant. The expression for the derivative of \( a^y \) (where \( a \) is a constant) with respect to \( y \) is \( a^y \cdot \ln(a) \). Thus, the partial derivative is: \[ \frac{\partial f}{\partial y} = x^y \cdot \ln(x) \]

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of calculus where functions depend on multiple variables, such as in the function \( f(x, y) = x^y \). It is essential for solving problems in higher dimensions, where variables interact in complex ways.
In multivariable calculus, we explore how the value of a function changes as each of its input variables changes, while keeping other variables constant. This leads us to the concept of partial derivatives.
  • These derivatives help us understand the slope or rate of change of the function in different directions.
  • Such an approach is crucial in fields like physics, engineering, and economics where multidimensional models are common.
Understanding partial derivatives empowers us to analyze and optimize these functions in real-world scenarios. By breaking down the changes due to each individual variable, we can make informed decisions based on this complex model.
Partial Derivative with Respect to x
The partial derivative with respect to \( x \) in multivariable calculus is found by treating all other variables as constants.
For the function \( f(x, y) = x^y \), finding the partial derivative with respect to \( x \) involves keeping \( y \) constant and differentiating as though \( x \) were the only variable.
  • The result is expressed as \( \frac{\partial f}{\partial x} = y \cdot x^{y-1} \).
  • This representation shows how \( f(x, y) \) changes along the \( x \)-direction, holding \( y \) fixed.
To achieve this, recall the power rule from single-variable calculus, applied here with \( y \) constant.
This type of differentiation allows us to explore the role of \( x \) independently in how the function behaves.
Partial Derivative with Respect to y
The partial derivative with respect to \( y \) is calculated by treating \( x \) as a constant. This approach helps to isolate the effect of changes in \( y \) on the function \( f(x, y) = x^y \).
Applying this to the given function, we derive \( \frac{\partial f}{\partial y} = x^y \cdot \ln(x) \).
  • Here, \( x \) is considered constant, and differentiation is with respect to \( y \).
  • Using the log function \( \ln(x) \) arises from the differentiation of an exponential function with a variable exponent.
This derivative tells us how the function grows or shrinks as \( y \) varies, with \( x \) unchanged.
Analyzing partial derivatives in different dimensions provides a deeper insight into function dynamics, critical in many scientific and engineering fields.

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Most popular questions from this chapter

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y, z, t)=x+y+z+t ; \quad x^{2}+y^{2}+z^{2}+t^{2}=1 $$

Find the extreme values of \(f\) on the region described by the inequality. $$ f(x, y)=e^{-x y}, \quad x^{2}+4 y^{2} \leq 1 $$

(a) Two surfaces are called orthogonal at a point of inter- section if their normal lines are perpendicular at that point. Show that surfaces with equations \(F(x, y, z)=0\) and \(G(x, y, z)=0\) are orthogonal at a point \(P\) where \(\nabla F \neq 0\) and \(\nabla G \neq 0\) if and only if $$F_{x} G_{x}+F_{y} G_{y}+F_{z} G_{z}=0$$ at \(P\) (b) Use part (a) to show that the surfaces $$z^{2}=x^{2}+y^{2}$$ and $$x^{2}+y^{2}+z^{2}=r^{2}$$ are orthogonal at every point of intersection. Can you see why this is true without using calculus?

The total production \(P\) of a certain product depends on the amount \(L\) of labor used and the amount \(K\) of capital investment. The Cobb-Douglas model for the production function is \(P=b L^{\alpha} K^{1-\alpha},\) where \(b\) and \(\alpha\) are positive constants and \(\alpha<1 .\) If the cost of a unit of labor is \(m\) and the cost of a unit of capital is \(n,\) and the company can spend only \(p\) dollars as its total budget, then maximizing the production \(P\) is subject to the constraint \(m L+n K=p\) . Show that the maximum production occurs when $$ L=\frac{\alpha p}{m} \quad \text { and } \quad K=\frac{(1-\alpha) p}{n} $$

Consider the problem of minimizing the function \(f(x, y)=x\) on the curve \(y^{2}+x^{4}-x^{3}=0\) (a piriform). (a) Try using Lagrange multipliers to solve the problem. (b) Show that the minimum value is \(f(0,0)=0\) but the Lagrange condition \(\nabla f(0,0)=\lambda \nabla g(0,0)\) is not satisfied for any value of \(\lambda\) . (c) Explain why Lagrange multipliers fail to find the minimum value in this case.
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