Question

Prove (Theorem 2.3) that \(\frac{d}{d x}\left[x^{n}\right]=n x^{n-1}\) for the case in which \(n\) is a rational number. (Hint: Write \(y=x^{p / q}\) in the form \(y^{q}=x^{p}\) and differentiate implicitly. Assume that \(p\) and \(q\) are integers, where \(q>0 .)\)

Short answer

The proof of the derivative of \(x^n\) where \(n\) is a rational number is \(\frac{dy}{dx} = n x^{n-1}\), done by using implicit differentiation, the chain rule and basic algebra.

Step-by-step solution

Write the Power as a Rational Number

Given a rational power \(p / q\) where \(p\) and \(q\) are integers and \(q > 0\), we can write the function \(y = x^{p/q}\) in the form \(y^q = x^p\). This step is crucial in allowing us to apply implicit differentiation.

Use Implicit Differentiation

Differentiate both sides of the equation with respect to \(x\), employing the chain rule. After differentiating we get \(q \cdot y^{q-1} \cdot \frac{dy}{dx} = p \cdot x^{p-1}\).

Simplify the Equation

We now solve the equation for the derivative. We divide both sides by \(q \cdot y^{q-1}\) leading to \(\frac{dy}{dx} = \frac{p}{q} \cdot \frac{x^{p-1}}{y^{q-1}}\).

Express \(y\) in terms of \(x\)

Substitute \(y = x^{p/q}\) back into the equation to express \(\frac{dy}{dx}\) solely in terms of \(x\). We get \(\frac{dy}{dx} = \frac{p}{q} \cdot \frac{x^{p-1}}{(x^{p/q})^{q-1}}\).

Simplify to get the Result

The expression \((x^{p/q})^{q-1}\) simplifies to \(x^{p-q}\), thus our equation becomes \(\frac{dy}{dx} = \frac{p}{q} x^{p-q}\). The expression \(p/q\) can be expressed as \(n\) and the expression \(p-q\) can be rearranged to \(n - 1\) or \(n -1/q\). This simplifies the equation to its final form: \(\frac{dy}{dx} = n x^{n-1}\).