Derivative Of 2x^6 Ln(13x^3 + 25x^2): Step-by-Step Solution

In this article, we'll walk through the process of finding the derivative of the function f(x) = 2x^6 ln(13x^3 + 25x^2). This involves applying the product rule and the chain rule, which are fundamental concepts in calculus. Derivatives are essential tools in mathematics, physics, engineering, and economics, used to model rates of change and optimize functions. Understanding derivatives helps in analyzing the behavior of functions, finding maximum and minimum values, and solving related rates problems. Before we dive into the solution, let's briefly review the necessary calculus rules.

Essential Calculus Rules

To find the derivative of the given function, we will primarily use two rules:

1. Product Rule: The product rule is used when differentiating a function that is the product of two other functions. If we have a function h(x) = u(x)v(x), then the derivative h'(x) is given by:

   h'(x) = u'(x)v(x) + u(x)v'(x)

   In simpler terms, the derivative of a product is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

2. Chain Rule: The chain rule is applied when differentiating a composite function. If we have a function g(x) = f(u(x)), then the derivative g'(x) is given by:

   g'(x) = f'(u(x)) * u'(x)

   This means the derivative of the composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Applying the Product Rule and Chain Rule: A Detailed Walkthrough

Now, let's apply these rules to find the derivative of f(x) = 2x^6 ln(13x^3 + 25x^2). This step-by-step solution will break down each part of the process, making it easier to understand. We will first identify the two functions being multiplied and then apply the product rule. Within the product rule, we'll need to use the chain rule to differentiate the natural logarithm part of the function. Let's begin by setting up our functions for the product rule.

Step 1: Identify u(x) and v(x)

In our function f(x) = 2x^6 ln(13x^3 + 25x^2), we can identify two main parts:

• u(x) = 2x^6
• v(x) = ln(13x^3 + 25x^2)

Now that we have identified u(x) and v(x), we can prepare to apply the product rule. The next step involves finding the derivatives of both u(x) and v(x). This is crucial for the product rule formula, which requires u'(x) and v'(x). We will start with the simpler function, u(x), and then move on to the more complex, v(x), which will require the chain rule.

Step 2: Find u'(x)

To find the derivative of u(x) = 2x^6, we use the power rule. The power rule states that if you have a function of the form x^n, its derivative is nx^(n-1). Applying this rule to u(x), we get:

• u'(x) = 2 * 6x^(6-1)
• u'(x) = 12x^5

So, the derivative of u(x) is 12x^5. Now, let's move on to finding the derivative of v(x), which is a bit more complex and requires the chain rule. Finding this derivative is a key step in solving the overall problem.

Step 3: Find v'(x)

To find the derivative of v(x) = ln(13x^3 + 25x^2), we need to apply the chain rule. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case:

• Outer function: ln(u)
• Inner function: u(x) = 13x^3 + 25x^2

The derivative of ln(u) with respect to u is 1/u. Now we need to find the derivative of the inner function, u(x) = 13x^3 + 25x^2.

• u'(x) = 13 * 3x^2 + 25 * 2x
• u'(x) = 39x^2 + 50x

Now, applying the chain rule, we get:

• v'(x) = (1 / (13x^3 + 25x^2)) * (39x^2 + 50x)
• v'(x) = (39x^2 + 50x) / (13x^3 + 25x^2)

We can simplify v'(x) by factoring out an x from both the numerator and the denominator:

• v'(x) = x(39x + 50) / x(13x^2 + 25x)
• v'(x) = (39x + 50) / (13x^2 + 25x)

Thus, the derivative of v(x) is (39x + 50) / (13x^2 + 25x). With both u'(x) and v'(x) found, we can now apply the product rule.

Step 4: Apply the Product Rule

Now that we have u(x) = 2x^6, u'(x) = 12x^5, v(x) = ln(13x^3 + 25x^2), and v'(x) = (39x + 50) / (13x^2 + 25x), we can apply the product rule:

f'(x) = u'(x)v(x) + u(x)v'(x)

Plugging in our functions and derivatives, we get:

f'(x) = (12x^5) * ln(13x^3 + 25x^2) + (2x^6) * ((39x + 50) / (13x^2 + 25x))

This is the derivative, but we can simplify it further to make it more readable. Simplifying this expression will give us a more concise answer.

Step 5: Simplify the Expression

To simplify the expression, we can first factor out common terms and then combine the terms:

f'(x) = 12x^5 ln(13x^3 + 25x^2) + (2x^6(39x + 50)) / (13x^2 + 25x)

Let's simplify the second term by factoring out x from the denominator:

f'(x) = 12x^5 ln(13x^3 + 25x^2) + (2x^6(39x + 50)) / (x(13x + 25))

Now, we can cancel out one x from the numerator and denominator:

f'(x) = 12x^5 ln(13x^3 + 25x^2) + (2x^5(39x + 50)) / (13x + 25)

To combine the terms, we need a common denominator. Multiply the first term by (13x + 25) / (13x + 25):

f'(x) = (12x^5 ln(13x^3 + 25x^2)(13x + 25)) / (13x + 25) + (2x^5(39x + 50)) / (13x + 25)

Now, we can combine the numerators:

f'(x) = (12x^5 ln(13x^3 + 25x^2)(13x + 25) + 2x^5(39x + 50)) / (13x + 25)

Factor out 2x^5 from the numerator:

f'(x) = (2x^5 [6 ln(13x^3 + 25x^2)(13x + 25) + (39x + 50)]) / (13x + 25)

So, the simplified derivative is:

f'(x) = (2x^5 [6(13x + 25)ln(13x^3 + 25x^2) + 39x + 50]) / (13x + 25)

Conclusion

The derivative of the function f(x) = 2x^6 ln(13x^3 + 25x^2) is:

f'(x) = (2x^5 [6(13x + 25)ln(13x^3 + 25x^2) + 39x + 50]) / (13x + 25)

We found this derivative by applying the product rule and the chain rule, along with careful simplification. This process illustrates the power and utility of calculus in analyzing complex functions. Understanding these rules is fundamental for further studies in mathematics and its applications.

For a deeper dive into calculus concepts, you might find resources on websites like Khan Academy Calculus helpful.