Chapter 13: Problem 51
Let \(f\) be a differentiable function of one variable, and let \(z=f(x+2 y) .\) Show that $$ 2 \frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}=0 $$
Short Answer
Expert verified
The equation holds true, as both sides simplify to zero.
Step by step solution
01
Identify the Composite Function
Given the function \( z = f(x + 2y) \), recognize that it is a composite function with an inner function \( u = x + 2y \). This is important for applying the chain rule.
02
Apply the Chain Rule for Partial Derivatives
To find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), use the chain rule. Since \( z = f(u) \), where \( u = x + 2y \), we have \( \frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} \) and \( \frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} \).
03
Compute \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\)
Calculate the partial derivatives of \( u \) with respect to \( x \) and \( y \). Since \( u = x + 2y \), we have \( \frac{\partial u}{\partial x} = 1 \) and \( \frac{\partial u}{\partial y} = 2 \).
04
Find \(\frac{\partial z}{\partial x}\)
Using the chain rule, \( \frac{\partial z}{\partial x} = f'(u) \cdot 1 = f'(x + 2y) \).
05
Find \(\frac{\partial z}{\partial y}\)
Similarly, \( \frac{\partial z}{\partial y} = f'(u) \cdot 2 = 2f'(x + 2y) \).
06
Set Up the Equation
Substitute the partial derivatives back into the equation \( 2 \frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = 0 \). We get \( 2f'(x + 2y) - 2f'(x + 2y) = 0 \).
07
Verify the Equation
Simplify the expression \( 2f'(x + 2y) - 2f'(x + 2y) \). Because both terms cancel each other, the result is \( 0 = 0 \), confirming the equation holds true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus that helps us differentiate composite functions. A composite function is formed when one function is nested inside another. For instance, if you have a function \( z = f(u) \) and the inner function is \( u = g(x, y) \), the chain rule allows you to differentiate \( z \) with respect to \( x \) and \( y \) by considering how \( z \) changes with \( u \), and how \( u \) changes with \( x \) and \( y \).
To apply the chain rule for partial derivatives, you need to:
The chain rule simplifies complex differentiation tasks, making it an indispensable tool in multivariable calculus.
To apply the chain rule for partial derivatives, you need to:
- Differentiate the outer function \( z = f(u) \) with respect to \( u \), giving \( f'(u) \).
- Differentiate the inner function \( u \) with respect to \( x \) and \( y \) separately.
- \( \frac{\partial z}{\partial x} = f'(u) \cdot \frac{\partial u}{\partial x} \)
- \( \frac{\partial z}{\partial y} = f'(u) \cdot \frac{\partial u}{\partial y} \)
The chain rule simplifies complex differentiation tasks, making it an indispensable tool in multivariable calculus.
Composite Function
A composite function involves plugging one function into another. In our problem, \( z = f(x + 2y) \), we identify it as a composite function where \( f \) is the outer function and \( x + 2y \) as the inner function.
A composite function requires careful differentiation because changes in the input of the inner function affect the output of the entire function. Here, the inner function \( u = x + 2y \) determines both the influence of \( x \) and \( y \) on \( z \).
Understanding composite functions is crucial because:
A composite function requires careful differentiation because changes in the input of the inner function affect the output of the entire function. Here, the inner function \( u = x + 2y \) determines both the influence of \( x \) and \( y \) on \( z \).
Understanding composite functions is crucial because:
- They often appear in real-world problems where one process depends on another.
- They require the use of the chain rule for differentiation.
Differentiable Function
A differentiable function is a function that is smooth and continuous, meaning it doesn't have any sharp turns or breaks. For a function to be differentiable, its derivative must exist at each point in its domain.
In this exercise, \( f \) is a differentiable function, which implies:
With differentiable functions, you can:
In this exercise, \( f \) is a differentiable function, which implies:
- The derivative \( f'(x) \) exists, allowing us to use it when applying the chain rule.
- The function behaves predictably, enabling the accurate calculation of the rate of change.
With differentiable functions, you can:
- Determine how small changes in input lead to changes in output.
- Predict the behavior of complex systems at a local level.