Problem 89
Question
In Section \(9.4\), we proved that $$ \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} $$ for the special case when \(n=2\). Use the chain rule to show that $$ \frac{d}{d x}\left(x^{1 / n}\right)=\frac{1}{n} x^{1 / n-1} $$ for any nonzero integer \(n\), assuming that \(f(x)=x^{1 / n}\) is differentiable. Hint: Let \(f(x)=x^{1 / n}\) so that \([f(x)]^{n}=x\). Differentiate both sides with respect to \(x\).
Step-by-Step Solution
Verified Answer
Using the chain rule, we defined \(f(x) = x^{1/n}\) and differentiated both sides with respect to \(x\). After applying the chain rule, substituting \(f(x)\), and isolating \(f'(x)\), we found that the derivative of \(x^{1/n}\) is \(\frac{1}{n}x^{1/n - 1}\) for any nonzero integer \(n\).
1Step 1: Define Function and Differentiate Both Sides
Define \(f(x) = x^{1/n}\), which implies \(f(x)^n = x\), since if we raise both sides to the power \(n\), we get x on the right-hand side.
Now differentiate both sides with respect to \(x\), applying the chain rule on the left-hand side:
\[
\frac{d}{dx}[f(x)^n] = \frac{d}{dx}(x)
\]
2Step 2: Apply Chain Rule
Using the chain rule on the left-hand side, we get:
\[
n[f(x)]^{n-1}\cdot f'(x) = 1
\]
In the above equation, we applied chain rule where chain rule states that if we have to take the derivative of the function in the form \((u(v))'\), where \(u = f(x)\) and \(v = n\), then it equals to \(u'(v)\cdot v'(x)\), or in simpler terms, derivative of the outer function times derivative of the inner function.
3Step 3: Substitute \(f(x) = x^{1/n}\)
We know that \(f(x) = x^{1/n}\), substitute it back into the equation:
\[
n[x^{1/n}]^{n-1}\cdot f'(x) = 1
\]
This simplifies further to
\[
n \cdot x^{1-1/n} \cdot f'(x) = 1
\]
4Step 4: Isolate \(f'(x)\)
Our primary goal here is to find \(f'(x)\), so we isolate \(f'(x)\) on one side of the equation to get:
\[
f'(x) = \frac{1}{n} \cdot x^{1/n - 1}
\]
This completes the proof. For any nonzero integer \(n\), the derivative of \(x^{1/n}\) is indeed \(\frac{1}{n} x^{1/n-1}\).
Key Concepts
Differential CalculusPower RuleDerivative of FunctionsMathematical Proofs
Differential Calculus
Differential calculus is a branch of mathematics that deals with the study of how things change and the rate at which they change. It focuses on the concept of the derivative, which measures the rate of change of a function with respect to one of its variables. This is essential in understanding how to maximize or minimize functions, and is used across various disciplines, from physics to economics. In the context of our exercise, differential calculus helps us analyze the behavior of power functions, particularly when we are dealing with fractional exponents, like in the function
Derivatives allow us to find slopes of tangent lines to curves and rates of change, making them crucial in predicting and understanding phenomena graphically and in real-life application.
f(x) = x^{1/n}.Derivatives allow us to find slopes of tangent lines to curves and rates of change, making them crucial in predicting and understanding phenomena graphically and in real-life application.
Power Rule
One of the fundamental rules in differential calculus is the power rule. This rule simplifies the process of taking the derivative of functions where the variable x is raised to a power. The rule states that the derivative of
In applying this rule, we significantly simplify the process of differentiation, illustrating the importance of knowing and applying mathematical rules correctly. The power rule is a prime example of the elegance and utility of calculus in mathematical problem-solving.
x^n, with respect to x, is n*x^(n-1). It's a straightforward way to handle otherwise complex calculations of derivatives and is particularly powerful because it applies to any real number n, whether it is positive, negative, an integer, or a fraction.In applying this rule, we significantly simplify the process of differentiation, illustrating the importance of knowing and applying mathematical rules correctly. The power rule is a prime example of the elegance and utility of calculus in mathematical problem-solving.
Derivative of Functions
The derivative of a function at a certain point characterizes the rate at which the function's value changes at that point. Essentially, to compute the derivative, one must understand how to apply specific guidelines to different kinds of functions. These guidelines include the power rule, product rule, quotient rule, and the chain rule.
For functions, such as
For functions, such as
f(x) = x^(1/n), which involve compositions of functions (in this case, a power function within a root function), we often employ the chain rule to find the derivative. The essential idea is to differentiate the 'outer' function and multiply it by the derivative of the 'inner' function. The process breaks down what might be a complex differentiation problem into simpler steps that are easier to manage.Mathematical Proofs
Mathematical proofs are logical arguments that demonstrate the truth of a mathematical statement. They are the core of mathematical rigor, as they validate theorems, which are statements that have been proven to be true. In calculus, proofs are used to establish the validity of rules for differentiation, among other things.
In the given exercise, we use a proof to show that the rule for differentiating powers of x also applies to fractional exponents by leveraging the chain rule. The proof carries out a logical sequence of operations starting with a known function and applying established differentiation rules to arrive at the derivative. Proofs not only confirm that our operations are valid but also enhance our deeper understanding of mathematical concepts and their interrelations.
In the given exercise, we use a proof to show that the rule for differentiating powers of x also applies to fractional exponents by leveraging the chain rule. The proof carries out a logical sequence of operations starting with a known function and applying established differentiation rules to arrive at the derivative. Proofs not only confirm that our operations are valid but also enhance our deeper understanding of mathematical concepts and their interrelations.
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