Problem 35 - Calculus for Biology and Medicine

Question

The following limit represents the derivative of a function \(f\) at the point \((a, f(a))\) : $$ \lim _{h \rightarrow 0} \frac{2(a+h)^{2}-2 a^{2}}{h} $$

Step-by-Step Solution

Answer

The derivative of the function \( f(x) = 2x^2 \) at \( x = a \) is \( 4a \).

Step 1: Identify the Expression

The given limit is \( \lim _{h \rightarrow 0} \frac{2(a+h)^{2}-2 a^{2}}{h} \). This expression is meant to represent the derivative of a function at a given point using the definition of the derivative.

Step 2: Recognize the Function

The expression inside the limit suggests the function \( f(x) = 2x^2 \) because the expression \( 2(a+h)^{2} - 2a^2 \) relates to the change in function values.

Step 3: Evaluate the Expansion

Expand \( (a+h)^2 \) to obtain \( a^2 + 2ah + h^2 \). Multiply throughout by 2 to get \( 2a^2 + 4ah + 2h^2 \).

Step 4: Simplify the Difference

Substitute the expansion into the limit expression to yield \( \frac{2a^2 + 4ah + 2h^2 - 2a^2}{h} \). This simplifies to \( \frac{4ah + 2h^2}{h} \).

Step 5: Simplify the Fraction

Factor out \( h \) from the numerator: \( \frac{h(4a + 2h)}{h} \). Cancel \( h \) from the numerator and denominator to get \( 4a + 2h \).

Step 6: Evaluate the Limit

Take the limit as \( h \rightarrow 0 \) of \( 4a + 2h \). The limit is \( 4a \) because as \( h \) approaches 0, the contribution of \( 2h \) vanishes.

Key Concepts

Limits

Limits are a foundational concept in calculus, allowing us to understand how functions behave as they get closer to a specific point. When you see \( \lim_{h \to 0} \), it means we are interested in what happens to our function as \( h \) approaches zero.
This type of limit is crucial for defining derivatives and continuous functions.

• For instance, in our original exercise, we use limits to find the rate of change of a function \( f(x) = 2x^2 \) as \( x \) approaches \( a \).
• The process involves observing how the function values change as \( h \) gets very small, essentially looking at the behavior of the function at an infinitesimally small distance.

Understanding limits help us comprehend how changes in \( x \), however minute, affect \( f(x) \). This ability to "zoom in" on a point is vital in ensuring we grasp how functions behave in calculus.

Calculus

Calculus is the branch of mathematics that deals with finding properties of derivatives and integrals of functions. It allows us to calculate the rate of change (differentiation) and the area under curves (integration).
In the context of our problem, calculus enables us to understand the concept of finding the derivative at a point \( a \).

• The exercise we are looking at is focused on calculating the derivative at a given point using limits.
• By evaluating how a function changes, we use one of the two core concepts of calculus: differentiation.

Calculus is instrumental in a wide array of fields, from physics to economics, wherever the notion of change is essential. It provides tools to model and solve problems involving continuous change.

Definition of Derivative

The derivative of a function at a given point represents the rate at which the function's value changes as its input changes. It can be viewed as the slope of the tangent line to the function's graph at that point.
The derivative is formally defined using limits:

• The derivative \( f'(a) \) at point \( a \) is given by \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).
• This expression calculates how much \( f(x) \) increases or decreases per unit change in \( x \), at the exact instant.

In our specific example, the use of the function \( f(x) = 2x^2 \) and its derivative calculation involves plugging into this definition. By expanding and simplifying \( 2(a+h)^2 - 2a^2 \), we find the derivative \( 4a \), which describes how sharply \( 2x^2 \) increases at any point \( a \). This process illustrates how derivatives describe instantaneous rates of change, foundational to understanding complex real-world phenomena.