Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Derivatives of functions are computed below. Identify the rule used in each step. In a few steps the rule is an algebraic rule of exponents and not a derivative rule. a. \(\left[5 t^{4}-7 e^{t}\right]^{\prime}\) \(\left[5 t^{4}\right]^{\prime}-\left[7 e^{t]^{\prime}}\right.\) \(5\left[t^{4}\right]^{\prime}-7\left[e^{t}\right]^{\prime}\) \(5 \times 4 t^{3}-7\left[e^{t}\right]^{\prime}\) \(5 \times 4 t^{3}-7 \times e^{t}\) \(\begin{array}{l}\text { b. } \quad\left[\left(1+e^{t}\right)^{8}\right]^{\prime} \\ & 8\left(1+e^{t}\right)^{7}\left[1+e^{t}\right]^{\prime} \\ & 8\left(1+e^{t}\right)^{7}\left([1]^{\prime}+\left[e^{t}\right]^{\prime}\right) \\\ & 8\left(1+e^{t}\right)^{7}\left(0+\left[e^{t}\right]^{\prime}\right) \\\ & 8\left(1+e^{t}\right)^{7}\left(0+e^{t}\right) \\ & 8 e^{t}\left(1+e^{t}\right)^{7}\end{array}\) c. \(\left[e^{3 t}\right]^{\prime}\) \(\left[\left(e^{t}\right)^{3}\right]^{\prime}\) \(3\left(e^{t}\right)^{2}\left[e^{t}\right]^{\prime}\) \(3\left(e^{t}\right)^{2} \times e^{t}\) \(3 e^{2 t} \times e^{t}\) \(3 e^{3 t}\)

Step-by-Step Solution

Answer

Parts (a), (b), and (c) use linearity, constant multiplication, power and exponential rules, and chain rule.

Step 1: Break down derivatives for a

Given: \( \left[5 t^{4}-7 e^{t}\right]^{\prime} \). Break it down using linearity: the derivative of a sum is the sum of the derivatives.Rule used: Linearity of derivatives.

Step 2: Apply constant multiplication rule for a

Separate constants from the functions: \( 5 \left[t^{4}\right]^{\prime} - 7 \left[e^{t}\right]^{\prime} \).Rule used: Constant multiplication rule.

Step 3: Differentiate power and exponential for a

Compute the derivatives: \( 5 \times 4 t^{3} - 7 \left[e^{t}\right]^{\prime} \) and \( 7 \times e^{t} \).Rules used: Power rule for \( t^{4} \) and exponential rule for \( e^{t} \).

Step 4: Simplify for a

Combine terms: \( 5 \times 4 t^{3} - 7 \times e^{t} \) gives the final derivative for part (a).No additional rules applied here, just algebraic simplification.

Step 1: Differentiate using chain rule for b

For \( \left[\left(1+e^{t}\right)^{8}\right]^{\prime} \), apply the chain rule to differentiate the composite function.Rule used: Chain rule.

Step 2: Inner function differentiation for b

Differentiate the inner function: \( 8 \left( 1 + e^{t} \right)^{7} \left( [1]^{\prime} + \left[e^{t}\right]^{\prime} \right) \).Rule used: Differentiate the sum inside the parentheses.

Step 3: Simplify inner derivative for b

\( 8 \left( 1 + e^{t} \right)^{7} \left( 0 + e^{t} \right) \). The derivative of a constant is zero.Rule used: Derivative of a constant.

Step 4: Final simplification for b

Multiply through: \( 8 e^{t} (1+e^{t})^{7} \). No additional rules applied here, just algebraic simplification.

Step 1: Chain rule application for c

For \( \left[e^{3 t}\right]^{\prime} \), rewrite it as: \( \left[\left(e^{t}\right)^{3}\right]^{\prime} \), then apply the chain rule to differentiate.Rule used: Chain rule.

Step 2: Differentiate the inner function for c

Compute \( 3\left(e^{t}\right)^{2} \left[e^{t}\right]^{\prime} \). The exponent is 3, so bring it down and reduce by 1.Rule used: Power rule.

Step 3: Simplification for c

Combine the differentiated terms: \( 3 \times e^{2t} \times e^{t} \).Rule used: Exponential rule.

Step 4: Combine and simplify for c

Finalize by combining for \( 3 e^{3 t} \).No additional rules applied here, just algebraic manipulation.

Key Concepts

Linearity of Derivatives

The linearity of derivatives is a fundamental concept in calculus. It tells us that the derivative of a sum is simply the sum of the derivatives. This also applies to the subtraction between functions. For example, when given the function \( [5t^4 - 7e^t]' \), we apply this rule by breaking it into two parts:

• the derivative of \( 5t^4 \)
• the derivative of \( -7e^t \)

By doing so, we can work on each component individually, simplifying our calculations. This principle makes it easier to handle complex expressions by tackling them in smaller, manageable pieces.

Power Rule

The power rule is a handy shortcut that makes finding derivatives straightforward when dealing with a power of a variable. The rule states that if you have a function of the form \( x^n \), its derivative is \( nx^{n-1} \). Let's illustrate this with an example:

• For \( t^4 \), applying the power rule gives \( 4t^{3} \).

This rule is particularly useful because it bypasses the need for lengthy computations and allows you to quickly determine the derivative of terms that have variables raised to powers. It's important to apply this rule carefully to avoid errors, especially with negative or fractional powers.

Chain Rule

The chain rule is essential for finding derivatives of composite functions, which are functions within other functions. It can be initially confusing but is straightforward once you get the hang of it. The chain rule states if you have a composite function \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \). Let’s see it in action:

• For the function \( (1+e^t)^8 \), the outer function is \( x^8 \) and the inner function is \( 1+e^t \).
• First, derive the outer function while keeping the inner function unchanged, giving \( 8(1+e^t)^7 \).
• Then, multiply this by the derivative of the inner function \( e^t \), hence the result: \( 8e^t(1+e^t)^7 \).

Using the chain rule, you can handle complex derivatives involving nested functions efficiently.

Constant Multiplication Rule

The constant multiplication rule simplifies the process of differentiation when a constant multiplies a function. It states that if \( c \) is a constant and \( f(x) \) is a differentiable function, then the derivative of \( cf(x) \) is \( c \cdot f'(x) \).For instance, in the expression \( 5t^4 \), we can separate the constant 5 from the variable part as we differentiate:

• The derivative is \( 5 \times 4t^3 \), where 4t^3 is the result of applying the power rule to \( t^4 \).

This rule is advantageous because it allows constants to remain unaffected during differentiation, significantly simplifying the differentiation process, especially in expressions with multiple terms.