Problem 24
Question
Use the Chain Rule-Power Rule to differentiate the given expression with respect to \(x\). $$ (x-1 / x)^{5 / 2} $$
Step-by-Step Solution
Verified Answer
The derivative of \((x-\frac{1}{x})^{\frac{5}{2}}\) is \(\frac{5}{2}(x - \frac{1}{x})^{\frac{3}{2}}(1 + \frac{1}{x^2})\).
1Step 1: Identify the Outer and Inner Functions
The given expression is \((x - \frac{1}{x})^{\frac{5}{2}}\). Here, the outer function is \(u^{\frac{5}{2}}\) where \(u = x - \frac{1}{x}\). This means that \(u\) serves as the inner function of the composite expression.
2Step 2: Differentiate the Outer Function Using the Power Rule
Differentiate \(u^{\frac{5}{2}}\) with respect to \(u\). The power rule states that if \(f(u) = u^n\), then \(f'(u) = nu^{n-1}\). Applying this, we have: \(\frac{d}{du}(u^{\frac{5}{2}}) = \frac{5}{2}u^{\frac{3}{2}}\).
3Step 3: Differentiate the Inner Function
Now differentiate the inner function \(u = x - \frac{1}{x}\) with respect to \(x\). The derivative of \(x\) is \(1\) and the derivative of \(-\frac{1}{x}\) is \(\frac{1}{x^2}\). Therefore, \(\frac{du}{dx} = 1 + \frac{1}{x^2}\).
4Step 4: Apply the Chain Rule
The Chain Rule states that \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). Substitute the derivatives from the previous steps: \(\frac{dy}{dx} = \frac{5}{2}(x - \frac{1}{x})^{\frac{3}{2}} \cdot (1 + \frac{1}{x^2})\).
5Step 5: Simplify the Expression
To simplify, distribute the term \(\frac{5}{2}(x - \frac{1}{x})^{\frac{3}{2}}\) across \((1 + \frac{1}{x^2})\). You can write the expression as: \[\frac{5}{2}(x - \frac{1}{x})^{\frac{3}{2}} \left(1 + \frac{1}{x^2}\right)\], which is the final differentiated form.
Key Concepts
Power RuleDifferentiationComposite FunctionCalculus
Power Rule
The Power Rule is one of the basic rules of differentiation. It gives us a simple method to find the derivative of a power of a variable. The rule states that if you have a function
In other words, to differentiate \(u^n\), you bring down the exponent as a coefficient and reduce the exponent by one. For example, in the expression \((x - \frac{1}{x})^{\frac{5}{2}}\), we identified the outer function as \(u^{\frac{5}{2}}\). Applying the Power Rule gives us \(\frac{d}{du}(u^{\frac{5}{2}}) = \frac{5}{2}u^{\frac{3}{2}}\). This simple process makes the Power Rule an essential tool for calculus.
- \(f(u) = u^n\)
- \(f'(u) = nu^{n-1}\)
In other words, to differentiate \(u^n\), you bring down the exponent as a coefficient and reduce the exponent by one. For example, in the expression \((x - \frac{1}{x})^{\frac{5}{2}}\), we identified the outer function as \(u^{\frac{5}{2}}\). Applying the Power Rule gives us \(\frac{d}{du}(u^{\frac{5}{2}}) = \frac{5}{2}u^{\frac{3}{2}}\). This simple process makes the Power Rule an essential tool for calculus.
Differentiation
Differentiation refers to the process of finding the derivative of a function. The derivative shows us how a function changes as its input changes. In calculus, this process helps understand the behavior of functions.
The derivative of a function can tell us several things:
a. Differentiating the outer function using the Power Rule, and
b. Differentiating the inner function \(u = x - \frac{1}{x}\). We find that the derivative of \(x\) is 1 and for \(\frac{-1}{x}\), it is \(\frac{1}{x^2}\).
By combining these steps with the Chain Rule, we find the derivative of the composite function.
The derivative of a function can tell us several things:
- The rate of change of the function
- Where the function increases or decreases
- The presence and location of local maxima and minima
a. Differentiating the outer function using the Power Rule, and
b. Differentiating the inner function \(u = x - \frac{1}{x}\). We find that the derivative of \(x\) is 1 and for \(\frac{-1}{x}\), it is \(\frac{1}{x^2}\).
By combining these steps with the Chain Rule, we find the derivative of the composite function.
Composite Function
A composite function combines two or more functions, where the output of one function becomes the input to another. In mathematical terms, if you have two functions,
Understanding composite functions is crucial when differentiating expressions with nested functions. In our exercise, \((x - \frac{1}{x})^{\frac{5}{2}}\) is a perfect example. Here, the inner function \(u = x - \frac{1}{x}\) is plugged into the outer power function \(u^{\frac{5}{2}}\).
When differentiating composite functions, it's essential to apply the Chain Rule to efficiently tackle the problem. This will involve differentiating both the outer and the inner functions separately, then combining the results.
- \(f(x)\)
- \(g(x)\)
- \((f \circ g)(x) = f(g(x))\)
Understanding composite functions is crucial when differentiating expressions with nested functions. In our exercise, \((x - \frac{1}{x})^{\frac{5}{2}}\) is a perfect example. Here, the inner function \(u = x - \frac{1}{x}\) is plugged into the outer power function \(u^{\frac{5}{2}}\).
When differentiating composite functions, it's essential to apply the Chain Rule to efficiently tackle the problem. This will involve differentiating both the outer and the inner functions separately, then combining the results.
Calculus
Calculus is the branch of mathematics focused on rates of change and the accumulation of quantities. It is essential for understanding various phenomena in physics, engineering, and economics.
There are two main branches:
Utilizing rules such as the Power Rule and Chain Rule streamlines differentiation and helps us manage intricate expressions. This highlights the power and versatility of calculus in problem-solving.
There are two main branches:
- Differential Calculus: This deals with finding the rate of change or slopes of curves, expressed via derivatives.
- Integral Calculus: This looks at the accumulation of quantities, using integrals.
Utilizing rules such as the Power Rule and Chain Rule streamlines differentiation and helps us manage intricate expressions. This highlights the power and versatility of calculus in problem-solving.
Other exercises in this chapter
Problem 24
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