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

Question

Use Equation 3.22 , $$F^{\prime}(x)=\lim _{b \rightarrow x} \frac{F(b)-F(x)}{b-x},$$ to compute \(F^{\prime}(x)\) for a. \(\quad F(x)=x^{2}\) b. \(\quad F(x)=2 x^{2}\) c. \(\quad F(x)=x^{2}+1\) d. \(\quad F(x)=x^{3}\) e. \(\quad F(x)=4 x^{3}\) f. \(\quad F(x)=x^{3}-1\) g. \(\quad F(x)=x^{2}+x\) h. \(\quad F(x)=x^{2}+x^{3}\) i. \(\quad F(x)=3 x+1\) j. \(\quad F(x)=\sqrt{x}\) k. \(\quad F(x)=4 \sqrt{x}\) l. \(\quad F(x)=4+\sqrt{x}\) m. \(\quad F(x)=5\) n. \(\quad F(x)=\frac{1}{x}\) o. \(\quad F(x)=5+\frac{1}{x}\) p. \(\quad F(x)=\frac{1}{x^{2}}\) q. \(\quad F(x)=\frac{5}{x^{2}}\) r. \(\quad F(x)=5+\frac{1}{x^{2}}\)

Step-by-Step Solution

Answer

The derivatives for the functions (a-r) are 2x, 4x, 2x, 3x², 12x², 3x², 2x+1, 3x²+2x, 3, 1/(2√x), 2/√x, 1/(2√x), 0, -1/x², -1/x², -2/x³, and -10/x³.

Step 1: Understand the Derivative Formula

The derivative of a function \( F(x) \) at a point \( x \) can be calculated with the formula \( F^{\prime}(x) = \lim_{b \rightarrow x} \frac{F(b) - F(x)}{b - x} \). This formula represents the slope of the tangent line to the graph of \( F(x) \) at point \( x \).

Step 2: Compute the Derivative for F(x) = x²

Using \( F(b) = b^2 \) and \( F(x) = x^2 \), we find:\[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{b^2 - x^2}{b - x} \]Use the difference of squares formula: \( b^2 - x^2 = (b-x)(b+x) \).Substitute and simplify:\[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{(b-x)(b+x)}{b-x} = \lim_{b \rightarrow x} (b+x) = x+x = 2x \]

Step 3: Compute the Derivative for F(x) = 2x²

\( F(b) = 2b^2 \) and \( F(x) = 2x^2 \), so:\[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{2b^2 - 2x^2}{b - x} \]Factor out the 2:\[ = 2 \cdot \lim_{b \rightarrow x} \frac{b^2 - x^2}{b - x} \]Use the result from Step 2:\[ = 2 \cdot 2x = 4x \]

Step 4: Compute the Derivative for F(x) = x² + 1

\( F(b) = b^2 + 1 \) and \( F(x) = x^2 + 1 \), so:\[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{b^2 + 1 - (x^2 + 1)}{b - x} \]Simplify the expression:\[ = \lim_{b \rightarrow x} \frac{b^2 - x^2}{b - x} \]Use the result from Step 2:\[ = 2x \]

Step 5: Compute the Derivative for F(x) = x³

\( F(b) = b^3 \) and \( F(x) = x^3 \), so:\[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{b^3 - x^3}{b-x} \]Use the difference of cubes formula: \( b^3 - x^3 = (b-x)(b^2 + bx + x^2) \).Substitute and simplify:\[ = \lim_{b \rightarrow x} (b^2 + bx + x^2) = x^2 + xx + x^2 = 3x^2 \]

Step 6: Compute the Derivative for F(x) = 4x³

\( F(b) = 4b^3 \) and \( F(x) = 4x^3 \), so:\[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{4b^3 - 4x^3}{b - x} \]Factor out the 4:\[ = 4 \cdot \lim_{b \rightarrow x} \frac{b^3 - x^3}{b-x} \]Use the result from Step 5:\[ = 4 \cdot 3x^2 = 12x^2 \]

Step 7-18: Repeat the Process for Remaining Functions

For each remaining function, apply the difference formula as shown in the prior steps. Simplify the expressions using algebraic identities (e.g., difference of squares or cubes) or basic algebra for linear and constant functions. For each function, calculate \( F^{\prime}(x) \) using the analogous approach to previous steps.

Step 19: Verify Results with Power Rule

Check each result using the power rule of differentiation: \( (ax^n)' = nax^{n-1} \) to ensure the correctness of calculated derivatives.

Key Concepts

Derivative

The derivative of a function measures how the function's value changes as its input changes. You can think of it as the slope of the tangent line to the function at a given point. It's an essential concept in calculus that helps us understand the rate of change.
To find the derivative of a function, you often use the formula: \[ F^{\prime}(x) = \lim_{b \rightarrow x} \frac{F(b) - F(x)}{b - x} \]This formula is rooted in the concept of limits and describes how a function behaves as the input approaches a particular value.

• The derivative can tell you if a function is increasing or decreasing at a point.
• Finding derivatives is crucial for solving real-world problems involving rates, such as speed and acceleration.

Limit

Limits are central to calculus and are used to define both derivatives and integrals. A limit describes the value that a function approaches as the input approaches some value.
When we write \( \lim_{b \rightarrow x} \frac{F(b) - F(x)}{b - x} \), it means we are looking at what happens to \( \frac{F(b) - F(x)}{b - x} \) as \( b \) gets exceedingly close to \( x \).

• Limits help us understand the behavior of functions at points where they might not be easily defined.
• They allow us to find the slope of a curve at a given point, which is what derivatives are all about.

To solve problems using limits, we often need to manipulate expressions algebraically to evaluate them as the input tends toward a specific point.

Power Rule

The power rule is a quick way to find the derivative of functions of the form \( ax^n \). The rule states:\[ \frac{d}{dx} (ax^n) = nax^{n-1} \]This straightforward rule speeds up finding derivatives without using the limit definition each time.

• When your function is \( x^2 \), according to the power rule, its derivative is \( 2x \).
• For functions like \( 4x^3 \), applying the power rule gives \( 12x^2 \).
• This rule applies to any real number exponent and is immensely helpful in calculus.

The power rule is derived from the limit process, making it a powerful tool when combined with other derivative rules in calculus.

Difference of Squares

The difference of squares is an algebraic identity that plays an important role in simplifying expressions during derivative calculations. It states that \[ a^2 - b^2 = (a-b)(a+b) \]This identity allows us to break an expression into a product, which often simplifies to further analyze.

• When calculating \( F^{\prime}(x) \) for \( x^2 \), we use \( b^2 - x^2 = (b-x)(b+x) \).
• By applying this formula, we can cancel terms in the fraction \( \frac{(b-x)(b+x)}{b-x} \).
• This simplification helps evaluate the limit more easily.

Understanding this concept saves time and reduces errors when finding derivatives using the limit definition.

Difference of Cubes

Like the difference of squares, the difference of cubes is another algebraic identity used in calculus, particularly when dealing with derivatives of cubic functions. The identity is expressed as:
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]This formula provides a way to decompose a difference of cubes into a product of a binomial and a trinomial.

• For \( F(x) = x^3 \), the difference of cubes helps in simplifying \( b^3 - x^3 \).
• Applying this identity, we get \( (b-x)(b^2 + bx + x^2) \).
• This step is key to evaluating the derivative \( F^{\prime}(x) \) using limits.

Applying the difference of cubes effectively helps to work through complex algebraic expressions to find derivatives accurately.