Question

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) $$f(x)=x^{2} e^{3 x}$$

Short answer

Answer: The first derivative is \(f'(x) = e^{3x}(2x + 3x^2)\), the second derivative is \(f^{\prime \prime}(x) = e^{3x}(8x + 9x^2 + 2)\), and the third derivative is \(f^{\prime \prime \prime}(x) = e^{3x}(32x + 27x^2 + 14)\).

Step-by-step solution

Find \(f^\prime(x)\) using the product rule.

First, we will find the first derivative of \(f(x)=x^{2} e^{3 x}\). This function is the product of two functions, namely \(u(x)=x^2\) and \(v(x)=e^{3x}\). To find its derivative, we'll apply the product rule, which states that if \(y=u(x)v(x)\), then \(y'=u'(x)v(x)+u(x)v'(x)\). Let's differentiate \(u\) and \(v\) first: $$u'(x)=\dfrac{d}{dx} (x^{2}) = 2x$$ $$v'(x)=\dfrac{d}{dx} (e^{3x}) = 3e^{3x} \text{ (using the chain rule)}$$ Now we can calculate the first derivative using the product rule: $$f'(x) = u'(x)v(x) + u(x)v'(x) = 2x\cdot e^{3x} + x^2\cdot 3e^{3x}$$ Next, we'll simplify \(f'(x)\): $$f'(x) = 2xe^{3x} + 3x^2e^{3x} = e^{3x}(2x + 3x^2)$$

Find \(f^{\prime \prime}(x)\) using the product rule.

Now we'll find the second derivative (i.e., the derivative of \(f'(x)\)) using the product rule again. This time, let \(u(x)=e^{3x}\) and \(v(x)=2x + 3x^2\): $$u'(x)=\dfrac{d}{dx} (e^{3x}) = 3e^{3x}$$ $$v'(x)=\dfrac{d}{dx} (2x + 3x^2) = 2 + 6x$$ Now applying the product rule to find the second derivative: $$f^{\prime \prime}(x) = u'(x)v(x) + u(x)v'(x) = 3e^{3x}(2x + 3x^2) + e^{3x}(2 + 6x)$$ After simplifying, we have: $$f^{\prime \prime}(x) = 6xe^{3x} + 9x^2e^{3x} + 2e^{3x} + 6xe^{3x} = e^{3x}(8x + 9x^2 + 2)$$

Find \(f^{\prime \prime \prime}(x)\) using the product rule.

Finally, we'll find the third derivative (i.e., the derivative of \(f^{\prime \prime}(x)\)) using the product rule once more. Let \(u(x)=e^{3x}\) and \(v(x)=8x + 9x^2 + 2\): $$u'(x)=\dfrac{d}{dx} (e^{3x}) = 3e^{3x}$$ $$v'(x)=\dfrac{d}{dx} (8x + 9x^2 + 2) = 8 + 18x$$ Applying the product rule to find the third derivative: $$f^{\prime \prime \prime}(x) = u'(x)v(x) + u(x)v'(x) = 3e^{3x}(8x + 9x^2 + 2) + e^{3x}(8 + 18x)$$ Simplifying, we get: $$f^{\prime \prime \prime}(x) = 24xe^{3x} + 27x^2e^{3x} + 6e^{3x} + 8e^{3x} + 18xe^{3x} = e^{3x}(32x + 27x^2 + 14)$$ In summary, we have found the first three derivatives of the given function as follows: $$f'(x) = e^{3x}(2x + 3x^2)$$ $$f^{\prime \prime}(x) = e^{3x}(8x + 9x^2 + 2)$$ $$f^{\prime \prime \prime}(x) = e^{3x}(32x + 27x^2 + 14)$$

Key concepts

Product Rule

Understanding the product rule is essential when dealing with functions that are products of two or more functions, such as in our exercise where we have to differentiate the function f(x) = x^2 e^{3x}. In calculus, the product rule is a guidance to find the derivative of a product of two functions. It states that if we have a function y = u(x)v(x), then the derivative y' is given by y' = u'(x)v(x) + u(x)v'(x).

In the step-by-step solution, we see the product rule applied multiple times. First, to find the derivative f'(x), we differentiate both u(x) = x^2 and v(x) = e^{3x} separately, and then use the rule to combine them. Not only is the product rule used for finding the first derivative, but it is also applied subsequently for higher-order derivatives like f''(x) and f'''(x) in the solution provided.

The key to mastering the product rule is practice. It's important to be familiar with differentiating simple functions before trying to apply the product rule, as you'll need to differentiate the individual functions that make up the product.

Chain Rule

The chain rule is another fundamental component in calculus, especially when functions are composite, meaning that they involve another function nested inside. For example, when dealing with an exponential function like e^{3x}, we see a multiplication inside the exponent. The chain rule comes in handy to tackle this situation.

It states that to differentiate a composite function f(g(x)), you would take the derivative of the outer function f with respect to the inner function g, and then multiply it by the derivative of the inner function g with respect to x. Formally, if y = f(g(x)), the derivative y' is f'(g(x)) * g'(x).

In our given solution, we use the chain rule to differentiate v(x) = e^{3x}, considering e^{3x} as the outer function f and 3x as the inner function g. Thus, we find the derivative of f with respect to g, which is still e^{3x} since the derivative of e^x with respect to x is e^x. Then, we multiply it by the derivative of g, which is 3. The result is 3e^{3x}, as correctly shown in the step-by-step solution.

Higher-Order Derivatives

Higher-order derivatives are, as the name implies, derivatives beyond the first derivative. They represent the rate of change of rate of change, which can be an essential tool in various fields including physics, engineering, and economics to describe acceleration, curvature, and other complex phenomena.

In the provided exercise, we are asked to find not only the first derivative f'(x), but also the second f''(x) and third f'''(x) derivatives of the function f(x) = x^2 e^{3x}. Higher-order derivatives are computed by repeatedly applying derivative rules to the results obtained from the previous differentiation steps. The solution shows a sequential application of the product rule to obtain these successive derivatives.

It's crucial to keep track of each differentiation step carefully, as errors can compound quickly in higher-order derivatives. Moreover, simplifying after each differentiation step can significantly prevent errors and make the following steps clearer, as exemplified in the solution.

Exponential Functions

Exponential functions are a category of mathematical functions of the form f(x) = a^x, where a is a constant and x is the variable. The base a of the exponential function is a critical factor that determines its growth rate. These functions play a significant role in modeling growth and decay processes in various disciplines.

In our exercise, e^{3x} represents an exponential function where the base is the mathematical constant e, approximately equal to 2.71828. This particular base is vital in calculus due to its unique property where the rate of change — the derivative — of e^x is itself e^x. This property markedly simplifies calculations involving derivatives, as shown in the step-by-step solution provided.

When working on exercises involving exponential functions, understanding this property can save a lot of time and effort. Exponential functions can often seem complicated due to their rapid growth, but their derivatives follow elegant and predictable patterns, allowing for simpler manipulation when applying rules like the product and chain rule.