Question

In Exercises, find the second derivative. $$ f(x)=(3+2 x) e^{-3 x} $$

Short answer

The second derivative of the function \( f(x)=(3+2 x) e^{-3 x} \) is \( f''(x) = -(9x+13)e^{-3x} \)

Step-by-step solution

Compute the first derivative

The equation is a product of two functions, \( (3+2x) \) and \( e^{-3x} \). Apply the product rule of differentiation, which states that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second. The chain rule will be applied on \( e^{-3x} \) which gives \(-3e^{-3x}\). Therefore:\( f'(x) = (3+2x)' \cdot e^{-3x} + (3+2x) \cdot (e^{-3x})' \)\( f'(x) = 2e^{-3x} - 3(3+2x)e^{-3x} \)\( f'(x) = e^{-3x}(2-3(2x+3)) \)

Compute the second derivative

Now differentiate the first derivative to compute the second derivative, hence applying the product rule again.\( f''(x) = e^{-3x}'(2-3(2x+3)) + e^{-3x}(2-3(2x+3))' \)\( f''(x) = -3e^{-3x}(2-3(2x+3)) + e^{-3x}\cdot (-6) \)\( f''(x) = e^{-3x}(-9x-13) \)

Simplify

We can rewrite the second derivative to make it cleaner. The second derivative of the function will read as:\( f''(x) = -(9x+13)e^{-3x} \)

Key concepts

Product Rule of Differentiation

The product rule is a critical operation in calculus, particularly when faced with the task of differentiating the product of two functions. Imagine you're trying to find the rate at which two intersecting areas change as you adjust their dimensions. This scenario is precisely when the product rule comes into play.

Let's say we have two functions, represented by ‘u’ and ‘v.’ The product rule tells us that the derivative of their product ‘uv’ is given by the formula:
\[ (uv)' = u'v + uv' \]
Here, \( u' \) and \( v' \) represent the derivatives of ‘u’ and ‘v,’ respectively. In practical terms, we first differentiate ‘u’ while keeping ‘v’ as it is, then we do the opposite. Finally, we add the two results. This rule is essential when dealing with functions like \( (3+2x)e^{-3x} \), where each part of the product is a separate function of ‘x.’

Chain Rule

Like a magical key unlocking the derivatives of composite functions, the chain rule allows us to differentiate functions of a function. If you've ever dealt with nested dolls, then you can think of the chain rule as the process of revealing one doll inside another in sequence.

To apply the chain rule, consider a composite function represented as \( f(g(x)) \). The derivative is found using the formula:
\[ f'(g(x)) = f'(g) \times g'(x) \]
In simpler terms, differentiate the outer function ‘f’ with respect to the inner function ‘g’, and multiply it by the derivative of ‘g’ with respect to ‘x.’ It's a two-step dance: first, focus on the outer layer, then move inwards to the core.

Exponential Functions Differentiation

Understanding differentiation of exponential functions is akin to deciphering a secret growth code—from economics to science, this concept is everywhere. Exponential functions often appear in the form \( ae^{kx} \), where ‘a’ and ‘k’ are constants, and \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

When differentiating an exponential function like \( e^{kx} \), we find that the derivative is a proportion of the function itself, which is fairly unique among function types. Specifically, the derivative is:
\[ \frac{d}{dx}e^{kx} = ke^{kx} \]
This property reflects the exponential function's constant rate of growth relative to its value, a concept that's particularly important when studying natural growth processes or calculating compound interest.

Calculus Problem Solving

Solving calculus problems is like assembling a puzzle; it requires recognizing patterns and applying the correct principles. Once you've identified which rules apply, such as the product rule for differentiating products of functions or the chain rule for composite functions, the next steps follow a logical progression.

To effectively solve calculus problems, understanding the fundamental rules of differentiation is paramount. This includes knowing how to handle polynomial, exponential, and trigonometric functions, among others. Furthermore, simplifying expressions whenever possible can unveil simpler paths to the solution. In our subject exercise:
\[ f(x) = (3+2x)e^{-3x} \]
, we unfurl the layers of differentiation by strategically applying these rules. Initially, the product rule breaks down the function into manageable parts. Subsequently, the chain rule unwinds the derivative of the exponential component. By combining these methodologies, we unveil the function's rate of change at different points and its pattern of growth or decay over time.