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Chapter 14: Problem 36

$$ \begin{array}{l}{\text { Find } \partial w / \partial v \quad \text { when } \quad u=-1, v=2 \quad \text { if } \quad w=x y+\ln z} \\ {x=v^{2} / u, y=u+v, z=\cos u}\end{array} $$

Short Answer

Expert verified
The partial derivative \( \frac{\partial w}{\partial v} \) is \( -8 \).

Step by step solution

01

Express w in terms of v and u

Given the expression of \( w = xy + \ln z \), substitute \( x, y, \) and \( z \) with their respective expressions in terms of \( u \) and \( v \):\[w = \left(\frac{v^2}{u}\right)(u+v) + \ln(\cos u).\]
02

Simplify the expression for w

Simplify the expression obtained in Step 1:\[w = \frac{v^2(u+v)}{u} + \ln(\cos u) = v^2 + \frac{v^3}{u} + \ln(\cos u).\]
03

Differentiate with respect to v

Calculate the partial derivative \( \frac{\partial w}{\partial v} \):- Differentiate \( v^2 \) with respect to \( v \) to get \( 2v \).- Differentiate \( \frac{v^3}{u} \) with respect to \( v \) to get \( \frac{3v^2}{u} \).- \( \ln(\cos u) \) is constant with respect to \( v \), so its derivative is 0.Thus, \[\frac{\partial w}{\partial v} = 2v + \frac{3v^2}{u}.\]
04

Substitute the given values for u and v

Insert the values \( u = -1 \) and \( v = 2 \) into the expression:\[\frac{\partial w}{\partial v} = 2(2) + \frac{3(2)^2}{-1} = 4 - 12 = -8.\]
05

Finalize the result

The expression calculated with the substituted values gives the final answer for \( \frac{\partial w}{\partial v} \) when \( u = -1 \) and \( v = 2 \):\[\frac{\partial w}{\partial v} = -8.\]

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a powerful technique used when dealing with equations where the variables are not easily isolated. Instead of a function expressed explicitly as \( y = f(x) \), you often have an equation involving multiple variables, for example, implicitly defined as \( F(x,y) = 0 \). To find the derivative of one variable with respect to another, we apply differentiation to both sides of the equation:
  • The chain rule plays a crucial role. When differentiating a term containing multiple variables, you must account for each variable's contribution.
  • You differentiate each part concerning the desired variable while maintaining terms for others.
  • The result may contain implicit derivatives, like \( \frac{dy}{dx} \) when \( y \) cannot be easily separated from \( x \).
Using implicit differentiation ensures that you can extract the rates of change even when a system's relationships are buried beneath layers of interdependent variables.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of more than one variable. Unlike single-variable calculus, where functions \( y = f(x) \) exist on a line, multivariable functions \( f(x, y, z, ...) \) exist in a higher dimensional space. Here's what makes it unique:
  • Partial Derivatives: These are derivatives of multivariable functions taken with respect to one variable, treating other variables as constants. They represent the function's rate of change along one axis in a multidimensional space.
  • Gradient: A vector formed by the partial derivatives, pointing in the direction of the greatest rate of increase of the function, essential for optimization problems.
  • Higher Dimensions: Limits, continuity, and differentiability have higher-dimensional analogs, requiring new ways to assess them, like using directional derivatives.
Multivariable calculus is integral for understanding systems where changes occur in multiple directions, such as in physics and engineering.
Function of Several Variables
A function of several variables involves two or more independent inputs and produces an output dependent on those inputs. Functions like \( f(x, y, z) = xy + \ln(z) \) combine these inputs into outputs through formulas. Key points include:
  • Domain: It's crucial to determine where the function is defined. In the example, \( z = \cos(u) \) ensures \( u \) produces a valid cosine value.
  • Partial Derivatives: These allow us to examine how one input affects the function's value while holding the others constant, as demonstrated in the derivative \( \frac{\partial w}{\partial v} \).
  • Interdependencies: The variables can interact in complex ways, affecting how we perform operations like integration or differentiation.
Functions of several variables are foundational in fields like economics, data science, and any domain that models systems with multiple influencing factors.

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Most popular questions from this chapter

In Exercises \(31-38,\) find the absolute maxima and minima of the functions on the given domains. \(f(x, y)=4 x-8 x y+2 y+1\) on the triangular plate bounded by the lines \(x=0, y=0, x+y=1\) in the first quadrant

In Exercises \(55-60,\) verify that \(w_{x y}=w_{y x}\). $$\omega=\frac{3 x-y}{x+y}$$

Does a function \(f(x, y)\) with continuous first partial derivatives throughout an open region \(R\) have to be continuous on \(R ?\) Give reasons for your answer.

The discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) is zero at the origin for each of the following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimum, or neither at the origin by imagining what the surface \(z=f(x, y)\) looks like. Describe your reasoning in each case. $$ \begin{array}{ll}{\text { a. } f(x, y)=x^{2} y^{2}} & {\text { b. } f(x, y)=1-x^{2} y^{2}} \\ {\text { c. }} {f(x, y)=x y^{2}} & {\text { d. } f(x, y)=x^{3} y^{2}} \\ {\text { e. }} {f(x, y)=x^{3} y^{3}} & {\text { f. } f(x, y)=x^{4} y^{4}}\end{array} $$

If a function \(f(x, y)\) has continuous second partial derivatives throughout an open region \(R,\) must the first-order partial derivatives of \(f\) be continuous on \(R ?\) Give reasons for your answer.
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