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In Exercises \(1-22,\) find \(\partial f / \partial x\) and \(\partial f / \partial y\) $$f(x, y)=x^{y}$$

Short Answer

Expert verified
\( \frac{\partial f}{\partial x} = y \cdot x^{y-1}, \frac{\partial f}{\partial y} = x^y \ln(x) \).

Step by step solution

01

Compute Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \) where \( f(x, y) = x^y \), treat \( y \) as a constant. Apply the power rule for differentiation. The derivative is \( y \cdot x^{y-1} \) because \( x^y \) with respect to \( x \) is essentially \( y \cdot x^{y-1} \). Thus, \( \frac{\partial f}{\partial x} = y \cdot x^{y-1} \).
02

Compute Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \) where \( f(x, y) = x^y \), treat \( x \) as a constant. Use the identity and logarithmic differentiation method. Rewrite \( f(x, y) = e^{y \ln(x)} \) then differentiate \( e^{y \ln(x)} \) with respect to \( y \):\[ \frac{\partial f}{\partial y} = e^{y \ln(x)} \cdot \ln(x) = x^y \ln(x) \] Therefore, \( \frac{\partial f}{\partial y} = x^y \ln(x) \).

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Key Concepts

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

Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions with more than one variable. In our exercise, the function is defined as \( f(x, y) = x^y \), which depends on both \( x \) and \( y \). This means that instead of simple derivatives, we calculate partial derivatives.

Understanding multivariable calculus is crucial when dealing with functions of two or more variables, which commonly appear in fields like physics, economics, and engineering. When computing partial derivatives, you treat the variable you're differentiating with respect to as a variable, while treating all other variables as constants. This kind of differentiation allows you to analyze how changes in one variable affect a function, while keeping other variables constant.
Differentiation
Differentiation involves finding the derivative of a function, which represents the rate of change of the function with respect to one of its variables. For our function, \( f(x, y) = x^y \), this rate of change can be determined with respect to \( x \) or \( y \).To find the partial derivative with respect to \( x \), we treat \( y \) as a constant and apply differentiation rules as we would in single-variable calculus.

When differentiating with respect to \( y \), we apply the principles of logarithmic differentiation and use the property \( x^y = e^{y \ln(x)} \). Differentiating logarithmic expressions can simplify the process and helps when dealing with exponents or complex functions. This approach is incredibly useful in calculus, particularly when expressions involve powers of variables.
Power Rule
The power rule is a fundamental tool in calculus used to find the derivative of a power function. It states that if you have a function of the form \( x^n \), the derivative is given by \( n \cdot x^{n-1} \).

For our example, \( f(x, y) = x^y \), when finding the partial derivative with respect to \( x \), the power rule allows us to easily compute \( \frac{\partial f}{\partial x} \). Even though \( y \) is a variable in the context of the whole function, during partial differentiation with respect to \( x \), it's a constant, so the rule applies directly: \( y \cdot x^{y-1} \). This simplicity and efficiency make the power rule widely applicable and a time-saver in calculus.
Logarithmic Differentiation
Logarithmic differentiation is a powerful technique for differentiating functions that are a product of variables or have variable exponents. For the function \( f(x, y) = x^y \), it helps to rewrite it in a logarithmic form: \( e^{y \ln(x)} \).

This transformation makes it easier to differentiate with respect to \( y \). By applying the chain rule to \( e^{y \ln(x)} \), we find the derivative to be \( x^y \ln(x) \). Logarithmic differentiation simplifies the process of finding derivatives that involve variables raised to powers, particularly when those powers are also variables. This approach can provide clarity and efficiency when managing complex differential equations.

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Most popular questions from this chapter

In Exercises \(71-76\) , you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$ \begin{array}{l}{f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16,-3 \leq x \leq 3} \\ {-6 \leq y \leq 6}\end{array} $$

Value of a \(2 \times 2\) determinant If \(|a|\) is much greater than \(|b|,|c|,\) and \(|d|,\) to which of \(a, b, c,\) and \(d\) is the value of the determinant $$ f(a, b, c, d)=\left| \begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right| $$ most sensitive? Give reasons for your answer.

Express \(v_{x}\) in terms of \(u\) and \(y\) if the equations \(x=v \ln u\) and \(y=u \ln v\) define \(u\) and \(v\) as functions of the independent variables \(x\) and \(y,\) and if \(v_{x}\) exists. (Hint: Differentiate both equations with respect to \(x\) and solve for \(v_{x}\) by eliminating \(u_{x}\).)

In Exercises \(31-38,\) find the absolute maxima and minima of the functions on the given domains. \(f(x, y)=48 x y-32 x^{3}-24 y^{2}\) on the rectangular plate \(0 \leq x \leq\) \(1,0 \leq y \leq 1\)

Which order of differentiation will calculate \(f_{x y}\) faster: \(x\) first or \(y\) first? Try to answer without writing anything down. \begin{equation}\begin{array}{ll}{\text { a. }} & {f(x, y)=x \sin y+e^{y}} \\\ {\text { b. }} & {f(x, y)=1 / x} \\ {\text { c. }} & {f(x, y)=y+(x / y)} \\\ {\text { d. }} & {f(x, y)=y+x^{2} y+4 y^{3}-\ln \left(y^{2}+1\right)} \\\ {\text { e. }} & {f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { f. }} & {f(x, y)=x \ln x y}\end{array}\end{equation}

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