Question

Compute the derivative of the given function. $$f(x)=\left(x^{2}+x\right)^{5}\left(3 x^{4}+2 x\right)^{3}$$

Short answer

The derivative is \( f'(x) = [5(x^2 + x)^4 (2x + 1)](3x^4 + 2x)^3 + (x^2 + x)^5 [3(3x^4 + 2x)^2 (12x^3 + 2)] \).

Step-by-step solution

Understand the Product Rule

The function given is in the form of a product of two functions: \( u(x) = (x^2 + x)^5 \) and \( v(x) = (3x^4 + 2x)^3 \). We apply the product rule for derivatives, which is \((uv)' = u'v + uv'\). This means we need to separately calculate the derivatives of \( u(x) \) and \( v(x) \).

Differentiate \( u(x) = (x^2 + x)^5 \)

To differentiate \( u(x) = (x^2 + x)^5 \), we need to use the chain rule. Let \( g(x) = x^2 + x \), so \( u(x) = [g(x)]^5 \). Then, by the chain rule \( u'(x) = 5[g(x)]^4 g'(x) \). First, differentiate \( g(x) = x^2 + x \) to get \( g'(x) = 2x + 1 \). Thus, \( u'(x) = 5(x^2 + x)^4 (2x + 1) \).

Differentiate \( v(x) = (3x^4 + 2x)^3 \)

Like in the previous step, we use the chain rule. Let \( h(x) = 3x^4 + 2x \), so \( v(x) = [h(x)]^3 \). By the chain rule, \( v'(x) = 3[h(x)]^2 h'(x) \). Differentiate \( h(x) = 3x^4 + 2x \) to get \( h'(x) = 12x^3 + 2 \). Thus, \( v'(x) = 3(3x^4 + 2x)^2 (12x^3 + 2) \).

Apply the Product Rule

Using the product rule, we have \( (uv)' = u'v + uv' \). Substitute \( u' \) and \( v' \) into this equation: \( f'(x) = [5(x^2 + x)^4 (2x + 1)](3x^4 + 2x)^3 + (x^2 + x)^5 [3(3x^4 + 2x)^2 (12x^3 + 2)] \).

Simplify if Necessary

You can expand and simplify the expression if needed, though for some problems keeping it factored may be preferable for better clarity or further calculations. The main task of computing the derivative is completed at this stage.

Key concepts

Understanding the Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of a product of two or more functions. When you have a function that is a combination of two different functions, say \( u(x) \) and \( v(x) \), you cannot simply take each derivative separately and multiply them. Instead, you must apply the product rule. The formula for the product rule is:

• \((uv)' = u'v + uv'\)

This means that you take the derivative of the first function \( u(x) \), multiply it by the second function \( v(x) \), and then add the first function \( u(x) \) multiplied by the derivative of the second function \( v(x) \).
In the given exercise, we use this rule to find the derivative of \( f(x) = (x^2 + x)^5 (3x^4 + 2x)^3 \). By identifying each part of the product as separate functions, we can tackle the problem more effectively.

Applying the Chain Rule

The chain rule is another vital tool in calculus for dealing with derivatives, especially when you're working with composite functions. A composite function is essentially a function within another function. In our exercise, both parts of the product, \( u(x) = (x^2 + x)^5 \) and \( v(x) = (3x^4 + 2x)^3 \), are examples of composite functions, where we have an outer function raised to a power and an inner function.

• To apply the chain rule, first identify the inner function \( g(x) \) and outer function \( f(g(x)) \).
• The chain rule formula is: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).

Let's break it down for \( u(x) \). Here, the inner function is \( g(x) = x^2 + x \) and the outer function is \( f(x) = x^5 \). Taking the derivative, we have \( u'(x) = 5(x^2 + x)^4 (2x + 1) \).
The same process is repeated for \( v(x) \) with \( h(x) = 3x^4 + 2x \), giving us \( v'(x) = 3(3x^4 + 2x)^2 (12x^3 + 2) \). The chain rule makes handling these nested functions straightforward by applying differentiation in layers.

Mastering Differentiation Techniques

Differentiation techniques extend beyond basic rules and allow us to differentiate even the most complex functions with ease. Mastery of these techniques equips you to solve a wide range of calculus problems more effortlessly.

• Start with understanding how to apply basic rules like the constant rule, power rule, and sum rule.
• Move on to the product and chain rules, which are especially useful for polynomial and composite functions.

In our example, we confidently use both the product rule and the chain rule to differentiate polynomials raised to a power. The ability to see a function in terms of its components lets you break down complex expressions into manageable parts.
These techniques are not only essential for solving homework exercises but also form the foundation for more advanced topics in calculus. By continuously practicing these differentiation techniques, students develop a deeper understanding and appreciation for calculus, ultimately enhancing their problem-solving skills.