Chapter 13: Problem 49
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 partial derivatives show that the equation holds: \(2\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = 0\).
Step by step solution
01
Understand the Problem
We need to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for the function \(z = f(x + 2y)\) and show that the equation \(2\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = 0\) holds.
02
Use Chain Rule for Partial Derivatives
To find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), use the chain rule for derivatives, since \(z\) is expressed as a function of \(u = x + 2y\).
03
Calculate Partial Derivative with Respect to x
Since \(u = x + 2y\), when differentiating \(z = f(u)\) with respect to \(x\), treat \(y\) as constant. Then, \(\frac{\partial z}{\partial x} = f'(u) \frac{\partial u}{\partial x}\). We have \(\frac{\partial u}{\partial x} = 1\), so \(\frac{\partial z}{\partial x} = f'(u) \times 1 = f'(u)\).
04
Calculate Partial Derivative with Respect to y
Treat \(x\) as constant when differentiating with respect to \(y\). Use \(\frac{\partial z}{\partial y} = f'(u) \frac{\partial u}{\partial y}\). Here, \(\frac{\partial u}{\partial y} = 2\), so \(\frac{\partial z}{\partial y} = f'(u) \times 2 = 2f'(u)\).
05
Substitute and Simplify the Expression
Substitute \(\frac{\partial z}{\partial x} = f'(u)\) and \(\frac{\partial z}{\partial y} = 2f'(u)\) into the equation to verify: \[2\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = 2f'(u) - 2f'(u) = 0.\] The expression is simplified to zero, which satisfies the given equation.
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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 used for finding the derivative of a composite function. When dealing with partial derivatives, the chain rule is equally essential. In the context of our problem, the function \( z = f(x + 2y) \) is dependent on \( u = x + 2y \), which in turn depends on both \( x \) and \( y \). This is why the chain rule is our tool of choice.
To find the partial derivative \( \frac{\partial z}{\partial x} \), the chain rule tells us to differentiate \( z \) with respect to \( u \) first, and then \( u \) with respect to \( x \) because \( u \) is a function of \( x \) as well. Similarly, for \( \frac{\partial z}{\partial y} \), we differentiate \( z \) with respect to \( u \) again but this time take the derivative of \( u \) with respect to \( y \).
This method allows us to handle derivatives in situations where direct differentiation isn't possible. By recognizing the relationship between \( z \), \( u \), \( x \), and \( y \), we effectively break down the differentiation process using the chain rule. This makes solving complex derivative problems much more manageable.
To find the partial derivative \( \frac{\partial z}{\partial x} \), the chain rule tells us to differentiate \( z \) with respect to \( u \) first, and then \( u \) with respect to \( x \) because \( u \) is a function of \( x \) as well. Similarly, for \( \frac{\partial z}{\partial y} \), we differentiate \( z \) with respect to \( u \) again but this time take the derivative of \( u \) with respect to \( y \).
This method allows us to handle derivatives in situations where direct differentiation isn't possible. By recognizing the relationship between \( z \), \( u \), \( x \), and \( y \), we effectively break down the differentiation process using the chain rule. This makes solving complex derivative problems much more manageable.
Differentiable Function
A differentiable function is one that has a derivative at every point in its domain. This ensures that the function can be smoothly graphed without any abrupt changes in slope or direction. In our exercise, \( f \), the function applied to \( x + 2y \), is given as differentiable, which is crucial.
There are a couple of things to remember about differentiable functions:
There are a couple of things to remember about differentiable functions:
- They must be continuous because discontinuities mean the function does not have a derivative at those points.
- A differentiable function is often smooth, with no sharp edges or corners in its graph.
Partial Derivative
Partial derivatives are a type of derivative where we focus on how a multi-variable function changes with respect to one variable, holding all other variables constant. It is a central concept in multivariable calculus and provides insights into the function's behavior with respect to individual arguments.
For instance, in the given function \( z = f(x + 2y) \), computing \( \frac{\partial z}{\partial x} \) involves determining how \( z \) changes as \( x \) varies, keeping \( y \) fixed. This partial derivative, found using the chain rule in our example, turns out to be \( f'(u) \).
Similarly, \( \frac{\partial z}{\partial y} \) involves observing the change in \( z \) as \( y \) changes, with \( x \) treated as constant. This is computed as \( 2f'(u) \) in the solution process. Partial derivatives help us dive deeper into understanding multi-variable scenarios by focusing on the change concerning individual components of the whole, thus uncovering intricate layers of function behavior.
For instance, in the given function \( z = f(x + 2y) \), computing \( \frac{\partial z}{\partial x} \) involves determining how \( z \) changes as \( x \) varies, keeping \( y \) fixed. This partial derivative, found using the chain rule in our example, turns out to be \( f'(u) \).
Similarly, \( \frac{\partial z}{\partial y} \) involves observing the change in \( z \) as \( y \) changes, with \( x \) treated as constant. This is computed as \( 2f'(u) \) in the solution process. Partial derivatives help us dive deeper into understanding multi-variable scenarios by focusing on the change concerning individual components of the whole, thus uncovering intricate layers of function behavior.