Unlocking Calculus: Differentiating A Complex Function

Alright, guys, let's dive into some calculus! We've got a function to differentiate, and it looks a bit intimidating at first glance, but don't worry, we'll break it down step by step. Our main goal here is to find the first, second, and third derivatives of the function and then evaluate the third derivative at a specific point. So, let's get started. The function we're dealing with is $f(x) = 71x^4 \ln(x) + 5e^{4x}$. This function is a combination of a product (the $71x^4 \ln(x)$ part) and an exponential term. Remember, the derivative of a function tells us the rate at which the function's output changes with respect to its input.

Before we start, let's refresh some key derivative rules. We'll need the product rule, the chain rule, and the basic derivatives of logarithmic and exponential functions. The product rule states that if we have a function $u(x)v(x)$, its derivative is $u'(x)v(x) + u(x)v'(x)$. The chain rule is used when we have a composite function, like $f(g(x))$. The derivative is $f'(g(x))g'(x)$. Also, recall that the derivative of $\ln(x)$ is $1/x$, and the derivative of $e^x$ is $e^x$. We're going to apply these to find our derivatives. In the world of calculus, differentiation is a fundamental concept, which provides a tool for determining how a function changes. The process involves finding the derivative, which represents the instantaneous rate of change of the function at any given point. To differentiate a function, one must apply the rules of differentiation, such as the product rule, chain rule, and power rule, depending on the structure of the function. For example, the power rule is used to differentiate terms with exponents, while the product rule is used for the product of two functions. These rules, when applied systematically, enable the determination of the derivative of complex functions. This understanding of differentiation is crucial in many areas, including physics, engineering, and economics, where it's used to model and analyze dynamic systems.

(a) Finding the First Derivative, $f'(x)$

First, let's find $f'(x)$. We have two terms in our function, so we'll differentiate them separately. For the first term, $71x^4 \ln(x)$, we'll need the product rule. Let $u(x) = 71x^4$ and $v(x) = \ln(x)$. Then, $u'(x) = 284x^3$ (using the power rule) and $v'(x) = 1/x$. Applying the product rule:

$(71x^4 \ln(x))' = u'(x)v(x) + u(x)v'(x) = 284x^3 \ln(x) + 71x^4(1/x) = 284x^3 \ln(x) + 71x^3$.

Now, let's differentiate the second term, $5e^{4x}$. We'll need the chain rule here. Let $g(x) = 4x$. Then, the derivative of $e^{4x}$ is $e^{4x} * 4 = 4e^{4x}$. So, the derivative of $5e^{4x}$ is $5 * 4e^{4x} = 20e^{4x}$.

Putting it all together, we get our first derivative:

$f'(x) = 284x^3 \ln(x) + 71x^3 + 20e^{4x}$.

This is the rate of change of the original function. It tells us how the output of the function changes as we change the input value, x. This first derivative is vital, as it sets the stage for further analysis, like finding critical points and understanding the function's increasing and decreasing intervals. The product rule and chain rule are essential tools in this context, allowing us to break down complex functions into more manageable parts. By understanding how to apply these rules, you can approach even the most complex differentiation problems with confidence. The derivative helps you understand how things change, which is a fundamental concept in many areas. Moreover, mastering differentiation techniques opens the door to deeper understanding of calculus concepts and their applications. It is important to remember that these mathematical tools are essential for the understanding of the function, and it is a key skill. Understanding and applying these rules accurately helps students develop strong analytical skills and a deeper understanding of mathematical concepts. The first derivative provides a complete view of how things change. Understanding the concepts of differentiation can open doors to exciting career opportunities.

(b) Calculating the Second Derivative, $f''(x)$

Next up, we need to find the second derivative, $f''(x)$. This is simply the derivative of $f'(x)$. So we need to differentiate $284x^3 \ln(x) + 71x^3 + 20e^{4x}$. Let's break it down again.

For the first term, $284x^3 \ln(x)$, we'll use the product rule again. Let $u(x) = 284x^3$ and $v(x) = \ln(x)$. Then $u'(x) = 852x^2$ and $v'(x) = 1/x$. Applying the product rule:

$(284x^3 \ln(x))' = u'(x)v(x) + u(x)v'(x) = 852x^2 \ln(x) + 284x^3(1/x) = 852x^2 \ln(x) + 284x^2$.

For the second term, $71x^3$, its derivative is $213x^2$ (using the power rule).

For the third term, $20e^{4x}$, its derivative is $80e^{4x}$ (using the chain rule).

Putting it all together, we get:

$f''(x) = 852x^2 \ln(x) + 284x^2 + 213x^2 + 80e^{4x} = 852x^2 \ln(x) + 497x^2 + 80e^{4x}$.

The second derivative tells us about the concavity of the original function. If $f''(x) > 0$, the function is concave up (like a smile), and if $f''(x) < 0$, the function is concave down (like a frown). This information is super important for understanding the overall shape of the graph of the function. Taking the second derivative is a step further into analyzing the function's behavior. The second derivative reveals critical information about the function's concavity, helping determine whether it curves upwards or downwards. This concept is fundamental to understanding the behavior and properties of a function, such as finding inflection points. Mastering the second derivative is crucial for various applications, including optimization problems, curve sketching, and understanding the rate of change of the rate of change. The ability to calculate the second derivative empowers students to perform comprehensive analysis, enhancing problem-solving capabilities in diverse fields.

(c) Finding the Third Derivative, $f'''(x)$

Now, let's find the third derivative, $f'''(x)$. This is the derivative of $f''(x)$. We need to differentiate $852x^2 \ln(x) + 497x^2 + 80e^{4x}$.

For the first term, $852x^2 \ln(x)$, we use the product rule. Let $u(x) = 852x^2$ and $v(x) = \ln(x)$. Then $u'(x) = 1704x$ and $v'(x) = 1/x$. Applying the product rule:

$(852x^2 \ln(x))' = u'(x)v(x) + u(x)v'(x) = 1704x \ln(x) + 852x^2(1/x) = 1704x \ln(x) + 852x$.

For the second term, $497x^2$, its derivative is $994x$.

For the third term, $80e^{4x}$, its derivative is $320e^{4x}$.

Putting it all together, we get:

$f'''(x) = 1704x \ln(x) + 852x + 994x + 320e^{4x} = 1704x \ln(x) + 1846x + 320e^{4x}$.

The third derivative provides deeper insights into the behavior of the function, especially in understanding its inflection points and the rate of change of concavity. It helps to analyze the function's behavior more precisely and identify critical points where the curvature changes. This process involves the application of differentiation rules to the second derivative, enabling us to uncover nuanced properties of the original function. The third derivative plays a significant role in advanced calculus, optimization problems, and curve analysis, offering a comprehensive understanding of the function's behavior. Understanding the third derivative is essential for a complete analysis of functions, including the identification of inflection points and the characterization of the function's concavity. Mastering the third derivative is key to advanced calculus, optimizing problems, and understanding the complete behavior of the function.

(d) Evaluating the Third Derivative at x = 2, $f'''(2)$

Finally, let's evaluate the third derivative at $x = 2$. We have $f'''(x) = 1704x \ln(x) + 1846x + 320e^{4x}$. So, we need to plug in $x = 2$:

$f'''(2) = 1704(2) \ln(2) + 1846(2) + 320e^{4(2)} = 3408 \ln(2) + 3692 + 320e^8$.

This is the exact form of the answer. If we want a decimal approximation, we calculate:

$3408 \ln(2) \approx 2367.66$

$320e^8 \approx 958611.75$

So, $f'''(2) \approx 2367.66 + 3692 + 958611.75 = 964671.41$.

Therefore, $f'''(2) \approx 964671.41$ (correct to 2 decimal places).

To recap, finding the derivatives involves understanding and applying the rules of differentiation, such as the product rule and chain rule, to break down complex functions into simpler components. The first derivative provides insight into the rate of change, the second derivative reveals concavity, and the third derivative offers a comprehensive understanding of the function's behavior. Evaluating the derivatives at specific points allows for a detailed analysis of the function's characteristics. This step-by-step process builds a foundation for a complete understanding of how calculus can be applied to diverse mathematical problems.

Conclusion:

In conclusion, we've successfully differentiated the given function step by step, finding its first, second, and third derivatives. We then evaluated the third derivative at $x = 2$. This process highlights the power of differentiation and how it helps us understand the behavior of functions. Keep practicing, and you'll become a differentiation pro in no time! Remember that understanding the product rule, the chain rule, and the basic derivatives of the main functions is essential. Also, it's important to be meticulous in each step to avoid errors. Always double-check your work! Great job, guys!