Problem 12 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let $$ P(t)=\frac{u(t)}{v(t)}=u(t) \times(v(t))^{-1} $$ Use the product rule and power chain rule to show that $$ P^{\prime}(t)=\frac{u^{\prime}(t) v(t)-u(t) v^{\prime}(t)}{v^{2}(t)} $$

Step-by-Step Solution

Answer

The derivative is \( P'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{v^2(t)} \).

Step 1: Product Rule Application

The given function is written as a product: \( P(t) = u(t) \times (v(t))^{-1} \). We need to find the derivative, \( P'(t) \), using the product rule. The product rule states that if you have two functions \( f(t) \) and \( g(t) \), \( (f \times g)' = f' g + f g' \). Apply this here with \( f(t) = u(t) \) and \( g(t) = (v(t))^{-1} \).

Step 2: Derivative of u(t)

Differentiate \( u(t) \) with respect to \( t \) to get \( u'(t) \).

Step 3: Derivative of (v(t))^{-1} using Power Chain Rule

Use the power rule for differentiation and the chain rule to differentiate \( (v(t))^{-1} \). The power rule states that \( (x^n)' = n \cdot x^{n-1} \). Applying it here, we start by rewriting \( (v(t))^{-1} \) as \( (v(t))^{-1} = (v(t))^{-1}\). Differentiating gives \((v(t))^{-1}' = -1 \times (v(t))^{-2} \times v'(t) \).

Step 4: Substitute Derivatives into Product Rule

Insert the derivatives from Step 2 and Step 3 into the product rule equation: \[ P'(t) = u'(t) \times (v(t))^{-1} + u(t) \times \left( -1 \times (v(t))^{-2} \times v'(t) \right) \]

Step 5: Simplify Expression

Simplify the expression obtained in Step 4:\[ P'(t) = \frac{u'(t)}{v(t)} - \frac{u(t) v'(t)}{v^2(t)} \]Combine the fractions:\[ P'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{v^2(t)} \]

Step 6: Conclusion

The expression \( P'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{v^2(t)} \) has been demonstrated using the product rule and chain rule. The quantity matches the standard formula for the derivative of a quotient.

Key Concepts

Product Rule

The Product Rule is essential in calculus when differentiating expressions that are the product of two separate functions. When you have a function like \( P(t) = u(t) \cdot v(t)^{-1} \), you apply the Product Rule to find its derivative. According to this rule:

• If you have two functions \( f(t) \) and \( g(t) \), their product's derivative is given by: \((f \cdot g)' = f' \cdot g + f \cdot g'\).

Here, we treat \( u(t) \) as \( f(t) \) and \( (v(t))^{-1} \) as \( g(t) \). Once you find the individual derivatives of these two parts, you plug them into this formula. This helps determine the change in the product of these two functions as it relates to a change in the variable, \( t \). The outcome equips you with a mechanism to handle more sophisticated functions with multiplication components.

Power Chain Rule

The Power Chain Rule combines elements of the power rule and the chain rule, and it's a powerhouse for differentiating composite functions. To employ this technique, look at functions of the form \( (v(t))^{-1} \).

• First, rewrite it as \((v(t))^{-n}\), where \( n=1\) for this exercise.
• Next, apply the power rule: \( (x^n)' = n \times x^{n-1}\).
• Finally, combine it with the chain rule, which requires you to multiply by the derivative of the inside function, \( v'(t) \).

Thus, the derivative of \((v(t))^{-1}\) becomes \(-1 \times (v(t))^{-2} \times v'(t)\). Through this approach, different layers of functions are handled systematically. It's about peeling back the outer functions until you can reach and differentiate the innermost one. This results in a nuanced understanding of how changes in input manifest through a series of layers.

Quotient Rule

The Quotient Rule is specifically designed to help with differentiating functions presented as one function divided by another. While the exercise reformulates the function into a product, understanding the connection to the Quotient Rule is still vital. The rule itself is beautifully efficient:

• For a function \( h(t) = \frac{u(t)}{v(t)} \), the derivative is \( h'(t) = \frac{u'(t) v(t) - u(t) v'(t)}{v^2(t)} \).

This formula is neatly derived from the Product and Chain Rules but deserves attention in its right. It handles each portion of the quotient and aligns them collectively in a single expression. The negatives and fraction division encapsulate how the slope of a quotient is impacted by both functions. Recognizing the equivalence between this and restructured Product Rule outcomes not only validates understanding but boosts confidence in handling these derivatives in various forms.