Problem 58 - Calculus Single Variable

Question

Let \(p\) and \(q\) be nonzero integers. Apply implicit differentiation to the equation \(x^{p}-y^{q}=0\) to obtain a proof of the Power Rule for rational exponents: $$ \frac{d}{d x}\left(x^{p / q}\right)=\frac{p}{q} \cdot x^{(p / q)-1} $$

Step-by-Step Solution

Answer

The implicit differentiation of \(x^p - y^q = 0\) results in \(\frac{dy}{dx} = \frac{p}{q} x^{(p/q)-1}\), proving the power rule for rational exponents.

Step 1: Understand the Given Equation

The given equation is \(x^p - y^q = 0\). We want to find \(\frac{dy}{dx}\) using implicit differentiation.

Step 2: Differentiate Both Sides with Respect to x

Differentiating both sides of \(x^p - y^q = 0\) concerning \(x\), we get:\[\frac{d}{dx}(x^p) - \frac{d}{dx}(y^q) = 0\]. Using differentiation rules, this becomes \[px^{p-1} - qy^{q-1}\cdot\frac{dy}{dx} = 0\].

Step 3: Solve for dy/dx

Rearranging \(px^{p-1} - qy^{q-1}\cdot\frac{dy}{dx} = 0\), we solve for \(\frac{dy}{dx}\) to obtain: \[qy^{q-1}\frac{dy}{dx} = px^{p-1}\]. Divide both sides by \(qy^{q-1}\) to find \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}}\].

Step 4: Substitute y with x^(p/q)

From the original equation, \(x^p = y^q\), we deduce \(y = x^{p/q}\). Substituting this into the equation for \(\frac{dy}{dx}\), we have \[\frac{dy}{dx} = \frac{px^{p-1}}{q(x^{p/q})^{q-1}}\].

Step 5: Simplify the Expression

Simplify \((x^{p/q})^{q-1}\) as \(x^{p-1}\) because \(y^q = x^p\) implies \(y = x^{p/q}\). Thus, \[\frac{px^{p-1}}{q x^{p-1}} = \frac{p}{q} \]. Since \(x=x\), the overall derivative simplifies to \[ \frac{p}{q} x^{p/q - 1} \,\] which is the power rule for rational exponents.

Key Concepts

Power Rule

The Power Rule is one of the fundamental differentiation techniques used in calculus. It provides an efficient way to differentiate functions of the form \( f(x) = x^n \), where \( n \) is any real number. The rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Simply, multiply the power by the coefficient in front of \( x \), and then subtract one from the power.

Using the power rule simplifies the differentiation process significantly, especially when dealing with simple polynomial functions. For example:

• If \( f(x) = x^3 \), then according to the power rule, \( f'(x) = 3x^{3-1} = 3x^2 \).
• For \( f(x) = 5x^4 \), using the power rule, the derivative \( f'(x) = 20x^3 \).

The beauty of this rule is in its straight-forward application, which will serve as a basis when understanding more complex differentiation techniques involving rational exponents.

Rational Exponents

Rational exponents extend the concept of powers to include fractions, providing a bridge between roots and exponents. When you have a rational exponent \( x^{p/q} \), it means:

• The base \( x \) is raised to a power \( p \),
• Then you take the \( q \)th root of it.

This concept allows us to express roots as exponents. For instance, \( x^{1/2} \) is equivalent to the square root of \( x \), and \( x^{3/2} \) stands for \( \sqrt{x^3} \). Rational exponents are vital in calculus since they enable integration and differentiation of root functions without switching formats.

When applying the power rule to rational exponents like \( x^{p/q} \), follow the same steps as for whole numbers: multiply by the exponent, then subtract one from it. Thus, the derivative of \( x^{p/q} \) is \( \frac{p}{q}x^{(p/q) - 1} \). This is precisely what the exercise covers, proving the power rule holds even when the exponents are fractions.

Differentiation Techniques

Differentiation techniques in calculus are tools that allow us to find derivatives of functions, revealing rates of change. Implicit differentiation, in particular, is a powerful method used when functions are not explicitly given as \( y = f(x) \).

With implicit differentiation, you differentiate both sides of an equation with respect to \( x \), treating \( y \) as an implicit function of \( x \). For example, given \( x^p - y^q = 0 \):

• Differentiate each term with respect to \( x \).
• Apply the chain rule where necessary, such as multiplying \( dy/dx \) when differentiating a term involving \( y \).
• Solve for \( dy/dx \) to find the derivative of \( y \) with respect to \( x \).

This technique is essential for finding derivatives of equations that cannot be rearranged into explicit functional forms easily. It broadens the ability to handle more complex expressions and proves indispensable when verifying derivative rules, such as the power rule for rational exponents.

Mastering implicit differentiation will equip you with the tools to tackle a wide range of calculus problems, enhancing both problem-solving skills and mathematical intuition.