Problem 64 $$\text {Find } f_{x x}, f_{x y}... [FREE SOLUTION]

Chapter 6: Problem 64

$$\text {Find } f_{x x}, f_{x y}, f_{y x}, \text { and } f_{y y}$$ $$f(x, y)=\frac{x y}{x-y}$$

Short Answer

Expert verified

The second-order partial derivatives are: $f_{xx} = \frac{2y^2}{(x-y)^3}$, $f_{xy} = -\frac{2y (x-2y)}{(x-y)^3}$, $f_{yx} = -\frac{2y (x-2y)}{(x-y)^3}$, and $f_{yy} = \frac{2x^2}{(x-y)^3}$.

Step by step solution

Compute the Partial Derivative with respect to x

First, find the first partial derivative of the function $f(x, y) = \frac{xy}{x - y}$ with respect to $x$. Use the quotient rule: \[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\]where $u = xy$ and $v = x - y$:\[\frac{\partial f}{\partial x} = \frac{(y \cdot (x-y)) - (xy \cdot 1)}{(x-y)^2} = \frac{y(x-y) - xy}{(x-y)^2} = \frac{yx - y^2 - xy}{(x-y)^2} = \frac{-y^2}{(x-y)^2}\]

Compute the Partial Derivative with respect to y

Now, find the first partial derivative of the same function with respect to $y$ using the quotient rule:\[\frac{\partial f}{\partial y} = \frac{(x \cdot (x-y)) - (xy \cdot (-1))}{(x-y)^2} = \frac{x(x-y) + xy}{(x-y)^2} = \frac{x^2 - xy + xy}{(x-y)^2} = \frac{x^2}{(x-y)^2}\]

Compute the Second Partial Derivative $f_{xx}$

Next, find the second partial derivative with respect to $x$. Compute the derivative of $f_x = -\frac{y^2}{(x - y)^2}$:Apply the chain rule and quotient rule again ($u = -y^2$ and $v = (x - y)^2$):\[\frac{d}{dx} \left( \frac{-y^2}{(x - y)^2} \right) = \frac{(0 \cdot (x - y)^2) - (-y^2 \cdot 2(x - y))}{((x - y)^2)^2} = \frac{2y^2(x - y)}{(x - y)^4} = \frac{2y^2}{(x-y)^3}\]

Compute the Second Partial Derivative $f_{xy}$

Compute the cross partial derivative with $f_{xy}$. Differentiate $f_x = -\frac{y^2}{(x - y)^2}$ with respect to $y$:Apply the quotient rule ($u = -y^2$ and $v = (x - y)^2$):\[\frac{d}{dy} \left( \frac{-y^2}{(x - y)^2} \right) = \frac{(-2y \cdot (x - y)^2) - (-y^2 \cdot 2(x - y))\cdot (-1)}{((x - y)^2)^2} = \frac{-2y (x-y)^2 + 2y^2 (x-y)}{(x-y)^4} = -\frac{2y (x-y) - 2y^2}{(x-y)^3}\]Simplify further:\[-\frac{2y (x-y) - 2y^2}{(x-y)^3} = -\frac{2yx - 2y^2 - 2y^2}{(x-y)^3} = -\frac{2yx - 4y^2}{(x-y)^3} = -\frac{2y (x-2y)}{(x-y)^3}\]

Compute the Second Partial Derivative $f_{yx}$

By Clairaut's theorem, if $f$ is continuous and all its second-order partial derivatives are continuous, then $f_{xy} = f_{yx}$. Therefore,\[ f_{yx} = -\frac{2y (x-2y)}{(x-y)^3} \]

Compute the Second Partial Derivative $f_{yy}$

Finally, find the second partial derivative with respect to $y$. Compute the derivative of $f_y = \frac{x^2}{(x - y)^2}$ with respect to $y$:Apply the quotient rule ($u = x^2$ and $v = (x - y)^2$):\[\frac{d}{dy} \left( \frac{x^2}{(x - y)^2} \right) = \frac{(0 \cdot (x - y)^2) - (x^2 \cdot 2(x - y))(-1)}{((x - y)^2)^2} = \frac{2x^2 (x - y)}{(x-y)^4} = \frac{2x^2}{(x-y)^3}\]

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Key Concepts

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

Partial Differentiation

Partial differentiation involves finding the derivative of a function with respect to one variable while keeping other variables constant. For a function like $f(x, y) = \frac{xy}{x-y}$, we can find the first partial derivatives $f_x$ and $f_y$. These derivatives help us understand how the function changes as each variable changes independently.

Quotient Rule

The quotient rule is used when differentiating a function that is the quotient of two other functions. The general form of the quotient rule is:\[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\]For example, in the case of $f(x, y) = \frac{xy}{x-y}$, when finding the partial derivatives $f_x$ and $f_y$, we treat the numerator $xy$ as $u$ and the denominator $x-y$ as $v$. Applying the quotient rule helps us break down the problem step by step.

Clairaut's Theorem

Clairaut's theorem states that if the mixed partial derivatives of a function are continuous, they are equal. In other words, $f_{xy} = f_{yx}$. For the function we're dealing with, $f(x, y) = \frac{xy}{x - y}$, we can use Clairaut's theorem to assert that the mixed partial derivatives are the same, simplifying our calculations. This holds true as long as the function's second-order partial derivatives are continuous, which they are in this example.

Chain Rule

The chain rule is a formula to compute the derivative of a composite function. It's used extensively when calculating second-order partial derivatives. For instance, when computing $f_{xx}$ or $f_{yy}$, the chain rule helps us handle the composition of multiple functions. In the function $f(x, y) = \frac{xy}{x - y}$, applying the chain rule allows us to manage derivatives of products and quotients more efficiently, ensuring we obtain accurate results through correct differentiation.

91影视

$$\text {Find } f_{x x}, f_{x y}, f_{y x}, \text { and } f_{y y}$$ $$f(x, y)=\frac{x y}{x-y}$$

Short Answer

Step by step solution

Compute the Partial Derivative with respect to x

Compute the Partial Derivative with respect to y

Compute the Second Partial Derivative \(f_{xx}\)

Compute the Second Partial Derivative \(f_{xy}\)

Compute the Second Partial Derivative \(f_{yx}\)

Compute the Second Partial Derivative \(f_{yy}\)

Key Concepts

Partial Differentiation

Quotient Rule

Clairaut's Theorem

Chain Rule

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