Problem 7 - The Calculus with Analytic Geometry

Question

Find the derivative of the given function. $$ h(u)=\left(3 u^{2}+5\right)^{3}(3 u-1)^{2} $$

Step-by-Step Solution

Answer

The derivative of the function is \(h'(u) = [18u(3u^2 + 5)^2][(3u - 1)^2] + [(3u^2 + 5)^3][6(3u - 1)]\).

Step 1: Identify the parts of the function

The function is a product of two functions: \(f(u) = \big(3u^2 + 5\big)^3\) and \(g(u) = \big(3u - 1\big)^2\).

Step 2: Apply the product rule

Use the product rule for differentiation: \(h'(u) = f'(u)g(u) + f(u)g'(u)\).

Step 3: Differentiate \(f(u)\)

To find \(f'(u)\), first use the chain rule. For \(f(u) = \big(3u^2 + 5\big)^3\), let \(v = 3u^2 + 5\). Then, \(f(u) = v^3\). The derivative with respect to \(u\) is \(f'(u) = 3v^2 \frac{dv}{du} = 3(3u^2 + 5)^2 \times 6u = 18u(3u^2 + 5)^2\).

Step 4: Differentiate \(g(u)\)

To find \(g'(u)\), use the chain rule again. For \(g(u) = \big(3u - 1\big)^2\), let \(w = 3u - 1\). Then, \(g(u) = w^2\). The derivative with respect to \(u\) is \(g'(u) = 2w \frac{dw}{du} = 2(3u - 1) \times 3 = 6(3u - 1) \).

Step 5: Combine the results

Substitute \(f(u), g(u), f'(u),\) and \(g'(u)\) into the product rule formula: \(h'(u) = [18u(3u^2 + 5)^2][(3u - 1)^2] + [(3u^2 + 5)^3][6(3u - 1)] \).

Step 6: Simplify the expression

Factor out common terms if possible to simplify.

Key Concepts

Product Rule

When we want to differentiate a function that is the product of two other functions, we employ the product rule. This rule simplifies finding the derivative when we have something like your given function:

• First function: \(f(u) = (3u^2 + 5)^3\)
• Second function: \(g(u) = (3u - 1)^2\)

The product rule states: \[h'(u) = f'(u)g(u) + f(u)g'(u)\]Let's break it down:

• First, we differentiate the first function, \(f(u)\), while keeping the second function, \(g(u)\), the same.
• Then, we differentiate the second function, \(g(u)\), while keeping the first function, \(f(u)\), the same.

Finally, we add these two results together to get our derivative.

Chain Rule

The chain rule helps us differentiate composite functions. In our exercise, both \(f(u)\) and \(g(u)\) are composite. That means we apply the chain rule to find their derivatives.

For \(f(u) = (3u^2 + 5)^3\), we set \(v = 3u^2 + 5\). Thus, \(f(u) = v^3\). The chain rule tells us:\[f'(u) = \frac{df}{dv} \frac{dv}{du} = 3v^2 \frac{dv}{du} = 3(3u^2 + 5)^2 \times 6u = 18u(3u^2 + 5)^2\]Similarly, for \(g(u) = (3u - 1)^2\), set \(w = 3u - 1\). Thus, \(g(u) = w^2\). The chain rule then tells us:\[g'(u) = \frac{dg}{dw} \frac{dw}{du} = 2w \frac{dw}{du} = 2(3u - 1) \times 3 = 6(3u - 1)\]

Differentiation Techniques

Differentiation is the process of finding a function's derivative. It involves various techniques like the product rule, chain rule, and others.Differentiation techniques are essential tools in calculus, helping us find the rate at which one quantity changes with respect to another. Here are some fundamental methods:

• Product Rule: Used when differentiating products of two functions. It states: \( (fg)' = f'g + fg' \).
• Chain Rule: Used for differentiating composite functions. It states: \( \frac{df}{dx} = \frac{df}{du} \frac{du}{dx} \).
• Power Rule: For a function \( x^n \), the derivative is: \ \frac{d}{dx} (x^n) = nx^{n-1} \.
• Quotient Rule: For a function that is a quotient of two functions. It's derivative formula is: \( \frac{d}{dx} \frac{f}{g} = \frac{f'g - fg'}{g^2} \).

Understanding and applying these techniques correctly makes solving complex differentiation problems more manageable.