Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose \(p\) and \(q\) are integers and \(u\) is a positive function that has a derivative at all numbers \(t\). Assume that \(\left[u^{\frac{p}{q}}(t)\right]^{\prime}\) exists. Give reasons for the steps \((i)-(i v)\) below that show $$ \left[u^{\frac{p}{q}}(t)\right]^{\prime}=\frac{p}{q} u^{\frac{p}{q}-1}(t) \times u^{\prime}(t) $$ Let $$ v(t)=u^{\frac{p}{q}}(t) $$ Then $$ \begin{aligned} v^{q}(t) &=u^{p}(t) \\ \left[v^{q}(t)\right]^{\prime} &=\left[u^{p}(t)\right]^{\prime} \\ q v^{q-1}(t) \times v^{\prime}(t) &=p u^{p-1}(t) \times u^{\prime}(t) \\ v^{\prime}(t) &=\frac{p u^{p-1}(t)}{q} \frac{p}{\left(u^{\frac{p}{q}}\right)^{q-1}} \times u^{\prime}(t) \\ \left[u^{\frac{p}{q}}(t)\right]^{\prime} &=\frac{p}{q} u^{\frac{p}{q}-1}(t) \times u^{\prime}(t) \end{aligned} $$

Step-by-Step Solution

Answer

Using substitution and differentiating using the chain and power rules, find \(\left[u^{\frac{p}{q}}(t)\right]^{\prime} = \frac{p}{q} u^{\frac{p}{q}-1}(t) \cdot u^{\prime}(t)\).

Step 1: Substitution

We start by setting \( v(t) = u^{\frac{p}{q}}(t) \). This simplifies the expressions we need to differentiate and analyze later.

Step 2: Power transformation on both sides

We transform the equation \( v^{q}(t) = u^{p}(t) \). This equation shows that raising \( v(t) \) to the power of \( q \) is equivalent to \( u(t) \) raised to the power of \( p \).

Step 3: Differentiate both sides using chain rule

Differentiating both sides with respect to \( t \), we invoke the chain rule: \( \left[v^{q}(t)\right]^{\prime} = \left[u^{p}(t)\right]^{\prime} \).

Step 4: Apply power rule and chain rule

On the left, the chain rule gives: \( q v^{q-1}(t) \cdot v^{\prime}(t) \). On the right, it results in: \( p u^{p-1}(t) \cdot u^{\prime}(t) \). Hence, the equation becomes: \( q v^{q-1}(t) \cdot v^{\prime}(t) = p u^{p-1}(t) \cdot u^{\prime}(t) \).

Step 5: Solve for derivative of \( v(t) \)

Rearranging the equation for \( v^{\prime}(t) \): \( v^{\prime}(t) = \frac{p u^{p-1}(t)}{q v^{q-1}(t)} \cdot u^{\prime}(t) \).

Step 6: Substitute back for \( v(t) \)

Since \( v(t) = u^{\frac{p}{q}}(t) \), \( v^{q-1}(t) = \left(u^{\frac{p}{q}}(t)\right)^{q-1} = u^{\frac{p-q}{q}}(t) \). Therefore, \( v^{\prime}(t) = \frac{p}{q} u^{\frac{p-q}{q}}(t) \cdot u^{\prime}(t) \).

Step 7: Express the final derivative

Thus, we conclude that \( \left[u^{\frac{p}{q}}(t)\right]^{\prime} = \frac{p}{q} u^{\frac{p}{q}-1}(t) \cdot u^{\prime}(t) \), which connects the derivative of the composition to the derivatives of \( u(t) \).

Key Concepts

Derivative

A derivative is a central concept in calculus that represents the rate at which a function changes. For any function, its derivative provides us with insight into how the function behaves at specific points.
In our exercise, we look at the function in the form of an exponent, specifically a fractional power. The derivative tells us how this function changes as its variable changes.
To find a derivative, we apply various rules like the Chain Rule and the Power Rule, which are valuable tools for dealing with complex functions. These tools simplify our calculations by decomposing complex expressions into manageable parts.

Power Rule

The Power Rule is a straightforward and frequently used rule in calculus that helps us to find the derivative of powers of a variable. For a function like \( f(x) = x^n \), where \( n \) is any real number, the derivative is given by \( f'(x) = n \cdot x^{n-1} \).
In this problem, the Power Rule is applied to functions with rational exponents, which are expressions like \( u^{\frac{p}{q}}(t) \). Even with fractional powers, the Power Rule holds, allowing us to simplify the derivative of the given function.
Using the Power Rule together with the Chain Rule allows us to express the derivative of a more complex function that includes inner functions—like \( u(t) \). This step is crucial in achieving the expression of the final derivative.

Functions

Functions are mathematical constructs that assign a unique output for every input, often expressed as \( f(x) \). They are the backbone of calculus and essential for understanding change and motion in mathematics.
In this exercise, \( u(t) \) represents a function, which is then raised to a fractional power to create a new function \( u^{\frac{p}{q}}(t) \).
Understanding how one function transforms into another through powers or compositions is key to unlocking their derivatives. In this process, substitutions can help simplify the structure of these functions, as demonstrated by introducing \( v(t) \) as a substitute for \( u^{\frac{p}{q}}(t) \).
Functions like these allow us to apply various rules, making the differentiation process more straightforward while ensuring that our final results remain accurate.

Integer Exponents

Integer exponents are numbers that describe how many times to multiply the base by itself. When discussing derivatives, integer exponents frequently appear as they simplify the initial understanding of differentiation.
In this exercise, powers are initially expressed with integer exponents (\( u^p(t) \)) alongside rational exponents. This approach illustrates the underlying structure of compound functions and how they can be broken down into simpler parts.
Combining integer and fractional exponents necessitates a careful application of both the Power Rule and the Chain Rule. As we solve for derivatives, we leverage these rules to handle the variety of exponents efficiently, delivering a result that adheres to calculus principles.