Problem 87 - Calculus and its applications

Question

Differentiate. $$ y=\sqrt[3]{x^{3}+6 x+1} \cdot x^{5} $$

Step-by-Step Solution

Answer

The derivative is \( y' = x^4 \left( (x^3 + 6x + 1)^{-2/3} (x^3 + 2x) + 5\sqrt[3]{x^3 + 6x + 1} \right) \).

Step 1: Identify the Product Rule

The function given is a product of two functions: \( u(x) = \sqrt[3]{x^3 + 6x + 1} \) and \( v(x) = x^5 \). To differentiate the product of two functions, use the product rule: \( (uv)' = u'v + uv' \).

Step 2: Differentiate \( u(x) = \sqrt[3]{x^3 + 6x + 1} \)

Rewrite \( u(x) \) as \( (x^3 + 6x + 1)^{1/3} \). Use the chain rule to differentiate: \( u'(x) = \frac{1}{3}(x^3 + 6x + 1)^{-2/3} \cdot (3x^2 + 6) \). Simplifying, \( u'(x) = (x^3 + 6x + 1)^{-2/3} \cdot (x^2 + 2) \).

Step 3: Differentiate \( v(x) = x^5 \)

The derivative of \( v(x) = x^5 \) is \( v'(x) = 5x^4 \) using the power rule.

Step 4: Apply the Product Rule

Using the product rule and substituting the derivatives obtained, the derivative of \( y \) is: \( y' = u'v + uv' = (x^3 + 6x + 1)^{-2/3} \cdot (x^2 + 2) \cdot x^5 + \sqrt[3]{x^3 + 6x + 1} \cdot 5x^4 \).

Step 5: Simplify the Expression

Simplify the expression by factoring out common terms: \( y' = x^4 \left( (x^3 + 6x + 1)^{-2/3} \, (x^2 + 2)x + 5\sqrt[3]{x^3 + 6x + 1} \right) \).

Key Concepts

Product Rule

In calculus differentiation, when we are dealing with the product of two functions, the product rule comes into play. The product rule is essential because it provides a straightforward method to differentiate the product of two functions. Here's how it works:

• Identify the two functions involved in the product. In our example, this is \( u(x) = \sqrt[3]{x^3 + 6x + 1} \) and \( v(x) = x^5 \).
• The product rule formula is \( (uv)' = u'v + uv' \).
• This means you differentiate the first function and multiply it by the second function. Then, add the product of the first function and the derivative of the second.

Understanding and applying this rule allows us to handle complex expressions efficiently. It unleashes the power of differentiating products without fuss.

Chain Rule

The chain rule is a vital concept in differentiation, especially when dealing with composite functions, like \( u(x) = (x^3 + 6x + 1)^{1/3} \). Let's break it down:

• First, identify the outer function and the inner function. Here, the outer function is \( f(z) = z^{1/3} \) and the inner function is \( z = x^3 + 6x + 1 \).
• The chain rule states: \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \).
• You start by differentiating the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function.

In our example, this results in \( \frac{1}{3}(x^3 + 6x + 1)^{-2/3} \cdot (3x^2 + 6) \). This technique helps navigate through layers of functions seamlessly.

Power Rule

The power rule is arguably one of the simplest differentiation rules and is used when differentiating functions of the form \( f(x) = x^n \). It's also crucial in our exercise.

• The power rule states: \( \frac{d}{dx} x^n = nx^{n-1} \).
• In simple words, multiply the function by the exponent and reduce the exponent by one.
• For example, differentiating \( v(x) = x^5 \) using the power rule gives us \( v'(x) = 5x^4 \).

This is an easy and direct way to find derivatives of polynomial functions, making it fundamental for learning calculus differentiation. Whether it's a straightforward function or part of a wider product, the power rule springs into action quickly.