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In Exercises \(63-66,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$f(x, y)=4+2 x-3 y-x y^{2}, \quad \frac{\partial f}{\partial x} \quad \text { and } \quad \frac{\partial f}{\partial y} \quad \text { at }(-2,1)$$

Short Answer

Expert verified
The partial derivatives are \( \frac{ abla f}{ abla x} = 2 \) and \( \frac{ abla f}{ abla y} = 1 \) at \((-2, 1)\).

Step by step solution

01

Understand the limit definition for partial derivatives in terms of x

To find the partial derivative of the function \(f(x, y)\) with respect to \(x\) at the point \((-2,1)\), we use the limit definition: \[ \frac{abla f}{abla x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h} \].This means we will vary \(x\) while keeping \(y\) constant.
02

Compute partial derivative with respect to x

Substitute \((x, y) = (-2, 1)\) into the limit definition:\[ \frac{abla f}{abla x}\bigg|_{(-2,1)} = \lim_{h \to 0} \frac{f(-2+h, 1) - f(-2, 1)}{h} \].First, find \(f(-2+h, 1) = 4 + 2(-2+h) - 3(1) - (-2+h)(1)^2 = 4 + 2h - 4 - 3 + 2 - h = 2h - 1\).Next, find \(f(-2, 1) = 4 + 2(-2) - 3(1) - (-2)(1)^2 = 4 - 4 - 3 + 2 = -1\).Now, compute the limit: \[ \frac{abla f}{abla x}\bigg|_{(-2,1)} = \lim_{h \to 0} \frac{2h - 1 + 1}{h} = \lim_{h \to 0} \frac{2h}{h} = 2 \]. Hence, the partial derivative with respect to \(x\) at the point \((-2, 1)\) is \(2\).
03

Understand the limit definition for partial derivatives in terms of y

Similarly, to find the partial derivative of \(f(x, y)\) with respect to \(y\) at the point \((-2, 1)\), we apply:\[ \frac{abla f}{abla y} = \lim_{k \to 0} \frac{f(x, y+k) - f(x, y)}{k} \].This means we will vary \(y\) while keeping \(x\) constant.
04

Compute partial derivative with respect to y

Substitute \((x, y) = (-2, 1)\) into the limit definition:\[ \frac{abla f}{abla y}\bigg|_{(-2,1)} = \lim_{k \to 0} \frac{f(-2, 1+k) - f(-2, 1)}{k} \].First, compute \(f(-2, 1+k) = 4 + 2(-2) - 3(1+k) - (-2)(1+k)^2 = 4 - 4 - 3 - 3k + 2(1 + 2k + k^2)\).Simplify: \(f(-2, 1+k) = -3 - 3k + 2 + 4k + 2k^2 = -1 + k + 2k^2\).Previously calculated, \(f(-2, 1) = -1\).Now, compute the limit:\[ \frac{abla f}{abla y}\bigg|_{(-2,1)} = \lim_{k \to 0} \frac{-1 + k + 2k^2 + 1}{k} = \lim_{k \to 0} \frac{k + 2k^2}{k} = \lim_{k \to 0} (1 + 2k) = 1 \].Hence, the partial derivative with respect to \(y\) at the point \((-2, 1)\) is \(1\).

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Key Concepts

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

Limit Definition
The limit definition is crucial for understanding the behavior of partial derivatives in multivariable calculus. When calculating the partial derivative, we focus on how the function changes by varying one variable while keeping others constant. This is mathematically expressed using limits. For instance, the partial derivative of a function with respect to a variable, such as \(x\), is given by: \[\frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}.\]This limit tells us how the function \(f(x, y)\) changes as \(x\) increases by a small amount \(h\), while keeping \(y\) unchanged. Understanding this approach gives us the necessary insight to evaluate the function's rate of change with respect to each variable.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. Examples include functions like \(f(x, y, z)\) or \(f(x, y)\). In multivariable calculus, each variable adds another dimension to the graph of the function, making it more complex to visualize and compute.
Handling multiple variables allows us to explore more intricate concepts, including gradients, divergent fields, and surface integrals.
Here, derivatives tell us how a function changes over different dimensions, which is crucial for applications in physics, engineering, and economics, where real-world scenarios often involve multiple factors.
Gradient
The gradient is a fundamental concept in multivariable calculus, often used to find the direction of greatest increase of a function. For a function of two variables \(f(x, y)\), the gradient is a vector that combines the partial derivatives with respect to \(x\) and \(y\): \[abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right).\]The gradient points in the direction where the function increases most rapidly, and its magnitude describes how fast the increase is.
This makes gradients essential for optimization problems and understanding the behavior of scalar fields.
Function of Two Variables
A function of two variables, noted as \(f(x, y)\), relates two independent variables \(x\) and \(y\) to a dependent variable \(z = f(x, y)\). It often represents surfaces in a three-dimensional space. Such functions are common in real-world applications like geographical maps and economic models.
When working with these functions, we evaluate how changes in \(x\) and \(y\) affect \(z\), which can be complex since changes in one variable might alter the dependent variable differently than the other variable does.
Partial derivatives are used to isolate the effect of changing just one input variable while keeping others constant, useful for studying such intricate functions.

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Most popular questions from this chapter

The Korteweg-deVries equation This nonlinear differential equation, which describes wave motion on shallow water surfaces, is given by $$4 u_{t}+u_{x x x}+12 u u_{x}=0.$$ Show that \(u(x, t)=\operatorname{sech}^{2}(x-t)\) satisfies the Kortweg-deVries equation.

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