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Magazine Chapter 3: Problem 38
Show that the power rule for derivatives applies to rational powers of the form \(y=x^{m / n}\) by raising both sides to the \(n^{\text {th }}\) power and using implicit differentiation.
Short Answer
Expert verified
The power rule applies: \( \frac{dy}{dx} = \frac{m}{n} x^{(m/n)-1} \).
Step by step solution
01
Raise Both Sides to the nth Power
Start by raising both sides of the equation \( y = x^{m/n} \) to the \( n^{\text{th}} \) power to eliminate the fractional exponent: \( y^n = (x^{m/n})^n \). This simplifies to \( y^n = x^m \).
02
Write the Implicit Differentiation Formulas
We'll differentiate both sides of the equation \( y^n = x^m \) with respect to \( x \). Recall that for implicit differentiation, you'll differentiate each term separately. For the left-hand side, use the chain rule: \( \frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx} \). For the right-hand side, use the power rule: \( \frac{d}{dx}(x^m) = mx^{m-1} \).
03
Differentiate Both Sides Using Implicit Differentiation
Differentiating both sides, we get: \( n y^{n-1} \frac{dy}{dx} = m x^{m-1} \). This equation relates the derivatives of \( y \) and \( x \).
04
Solve for the Derivative \( \frac{dy}{dx} \)
To isolate \( \frac{dy}{dx} \), divide both sides of the equation \( n y^{n-1} \frac{dy}{dx} = m x^{m-1} \) by \( n y^{n-1} \): \( \frac{dy}{dx} = \frac{m x^{m-1}}{n y^{n-1}} \).
05
Substitute \( y \) Back
Recall that \( y = x^{m/n} \), so \( y^{n-1} = (x^{m/n})^{n-1} = x^{m(n-1)/n} \). Substitute \( y^{n-1} \) back into the derivative equation: \( \frac{dy}{dx} = \frac{m x^{m-1}}{n x^{m(n-1)/n}} \).
06
Simplify the Expression
Simplify \( \frac{dy}{dx} = \frac{m x^{m-1}}{n x^{m(n-1)/n}} \) further by recognizing that the denominator is \( x^{m-1} \): \( \frac{dy}{dx} = \frac{m}{n} x^{(m-1) - m(n-1)/n} \). Simplifying the exponent gives \( \frac{dy}{dx} = \frac{m}{n} x^{(m/n)-1} \), confirming that the power rule applies.
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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 technique used when it's difficult or impossible to solve an equation explicitly for one variable in terms of another. In the given exercise, we are dealing with the equation \( y = x^{m/n} \) where the fractional exponent makes it challenging to directly find the derivative. Here's where implicit differentiation shines. To apply implicit differentiation:
- First, we raise both sides of the equation to the power of \( n \), resulting in \( y^n = x^m \). This step eliminates the fractional exponent and allows us to deal with a simpler form of the equation.
- Next, you differentiate both sides with respect to \( x \). On the left hand side, when differentiating \( y^n \), use the chain rule, giving \( \frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx} \).
- On the right side, apply the standard power rule for derivatives, \( \frac{d}{dx}(x^m) = mx^{m-1} \).
Rational Exponents
Rational exponents refer to exponents that are fractions or ratios of integers. In the expression \( x^{m/n} \), \( m/n \) is a rational exponent. These exponents extend the concept of exponents beyond integers and allow us to represent roots and powers in a unified notation. Here's how they work:
- The denominator \( n \) indicates the root of the base \( x \). For instance, \( x^{1/2} \) is the square root of \( x \).
- The numerator \( m \) represents the power to which the base is raised after taking the root.
- For example, \( x^{3/2} \) means we first find the square root of \( x \), then cube the result.
Derivative of Power Functions
The derivative of power functions is a core concept in calculus. Power functions are of the form \( x^a \), where \( a \) is any real number, including rational numbers. The power rule provides a simple formula to find derivatives of these functions: If \( f(x) = x^a \), then the derivative \( f'(x) = a \cdot x^{a-1} \). This rule applies to any kind of exponent, including rational ones.In our exercise:
- We first rewrite the function with rational exponents as \( y^n = x^m \) to facilitate differentiation.
- Taking derivatives, the power rule is used: For \( x^m \), it鈥檚 differentiated to \( mx^{m-1} \).
- This is contrasted with the implicit differentiation on the left side.
- Finally, when solving for \( \frac{dy}{dx} \), the power rule confirms its applicability to rational exponents: resulting in \( \frac{dy}{dx} = \frac{m}{n} x^{(m/n)-1} \).
