Question

Find the second derivative of the function. \(f(x)=(3+2 x) e^{-3 x}\)

Short answer

The second derivative of the function \(f(x) = (3+2x)e^{-3x}\) is \(f''(x) = 27xe^{-3x} + 3e^{-3x}\)

Step-by-step solution

First derivative

Apply the product rule: \(f'(x) = (3+2x)' \cdot e^{-3x} + (3+2x) \cdot (e^{-3x})'\). This simplifies to \(f'(x) = 2e^{-3x} + (-3)(3+2x)e^{-3x}\) after applying the chain rule.

Simplification

Simplify the expression of the first derivative: \(f'(x) = 2e^{-3x} - 9xe^{-3x} - 6e^{-3x}\). Combine the exponential terms, we have: \(f'(x) = -9xe^{-3x} - 4e^{-3x}\).

Second derivative

Repeat the first process for the second derivative, applying the product rule and then the chain rule. Here, \(g(x) = -9x\) or \(-4\) and \(h(x) = e^{-3x}\). Thus we get: \(f''(x) = (-9)'e^{-3x} + -9(e^{-3x})' - 4 \cdot e^{-3x}' \).

Simplification of second derivative

Next, simplify: \(f''(x) = -9e^{-3x} +27xe^{-3x} + 12e^{-3x}\). Combining the terms gives: \(f''(x) = 27xe^{-3x} + 3e^{-3x}\).

Key concepts

Product Rule

When taking the derivative of a function that is the product of two sub-functions, we apply the product rule. This is a cornerstone concept for calculus students, as many functions you'll encounter are products of simpler functions.

The product rule states that for two functions, say \( g(x) \) and \( h(x) \), the derivative of their product \( g(x) \cdot h(x) \) is \( g'(x) \cdot h(x) + g(x) \cdot h'(x) \). In the exercise, when finding the first derivative of \( f(x) = (3+2x)e^{-3x} \), we identified two sub-functions: \( g(x) = (3+2x) \) and \( h(x) = e^{-3x} \). Applying the product rule and then simplifying the resulting terms gave us the first derivative \( f'(x) \).

Chain Rule

The chain rule is used when differentiating a function composed of an outer function and an inner function; it's the derivative version of 'opening a nested doll'.

Formally, if \( f(x) = h(g(x)) \) where \( h \) and \( g \) are both differentiable, then the chain rule states \( f'(x) = h'(g(x)) \cdot g'(x) \). In the given function \( f(x) = (3+2x)e^{-3x} \) when differentiating the exponential part \( e^{-3x} \), we viewed \( e^{u} \) as the outer function and \( u = -3x \) as the inner function. The derivative of \( e^{u} \) with respect to \( u \) is \( e^{u} \) itself, and \( u' = -3 \) is the derivative of the inner function \( -3x \). This application ensures that every part of the function is correctly differentiated.

Exponential Functions Differentiation

Differentiating exponential functions follows specific rules, especially when the base of the exponential is \( e \), the mathematical constant approximately equal to 2.71828. The derivative of \( e^x \) with respect to \( x \) is simply \( e^x \), which is what makes \( e \) a unique and fundamental base for natural logarithms and calculus.

In our exercise, we dealt with the differential of \( e^{-3x} \). According to the chain rule, the derivative of \( e^{u} \) where \( u \) is a function of \( x \) is \( e^{u} \cdot u' \). So, the differentiation of \( e^{-3x} \) is \( e^{-3x} \cdot (-3) \). This step is crucial because it might seem counterintuitive that the function essentially reproduces itself during differentiation, but remembering this unique property of \( e^x \) is vital for calculus students.

Calculus Simplification Steps

Simplifying calculus expressions is an inevitable part of solving calculus problems, and it is a process where the main aim is to write complex expressions in a more concise form, making them easier to understand or further manipulate. After applying differentiation rules, we often end up with lengthy expressions that require combining like terms, factoring, or expanding to reach a more simplified result.

In the context of our problem, after finding the first derivative \( f'(x) \), we had to simplify by combining like terms. Similarly with the second derivative \( f''(x) \), we simplify the expression by combining the exponential terms. Understanding how to manipulate algebraic expressions and simplifying correctly are critical skills, because even if you have correctly applied the differentiation rules, an unsimplified answer can lead to confusion or the incorrect interpretation of the result.