To find the derivative of \((ax - k)^2 (bx - k)^2\), let’s go through the steps carefully. We’ll need to apply both the **product rule** and the **chain rule** in our derivation.

### Step 1: Define the Function and Simplify Notation
Let:
\[
f(x) = (ax - k)^2 (bx - k)^2
\]

Define \(u = (ax - k)^2\) and \(v = (bx - k)^2\), so:
\[
f(x) = u \cdot v
\]

### Step 2: Apply the Product Rule
The derivative of \(f(x) = u \cdot v\) with respect to \(x\) is:
\[
f'(x) = u' \cdot v + u \cdot v'
\]

Thus, we need to calculate \(u'\) and \(v'\).

### Step 3: Differentiate \(u = (ax - k)^2\)
Using the chain rule, we find \(u'\):
\[
u = (ax - k)^2
\]
\[
u' = 2(ax - k) \cdot \frac{d}{dx}(ax - k)
\]
The derivative of \(ax - k\) with respect to \(x\) is \(a\), so:
\[
u' = 2(ax - k) \cdot a = 2a(ax - k)
\]

### Step 4: Differentiate \(v = (bx - k)^2\)
Similarly, for \(v = (bx - k)^2\), we apply the chain rule:
\[
v' = 2(bx - k) \cdot \frac{d}{dx}(bx - k)
\]
The derivative of \(bx - k\) with respect to \(x\) is \(b\), so:
\[
v' = 2(bx - k) \cdot b = 2b(bx - k)
\]

### Step 5: Substitute \(u'\) and \(v'\) into the Product Rule
Now we have:
\[
f'(x) = u' \cdot v + u \cdot v'
\]
Substituting the values of \(u\), \(v\), \(u'\), and \(v'\):
\[
f'(x) = \left(2a(ax - k)\right)(bx - k)^2 + (ax - k)^2 \left(2b(bx - k)\right)
\]

### Step 6: Simplify the Expression
Let's factor out \(2\) to simplify:
\[
f'(x) = 2 \left[ a(ax - k)(bx - k)^2 + b(ax - k)^2(bx - k) \right]
\]

This is the derivative of \(f(x) = (ax - k)^2(bx - k)^2\).

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Would you like further simplification or details on any specific part of this derivation?

### Further Questions
1. How would you apply the product rule if more terms were multiplied?
2. What happens if \(a\) or \(b\) equals zero in this expression?
3. How would you find higher-order derivatives of this function?
4. What if we expanded \((ax - k)^2(bx - k)^2\) first and then differentiated?
5. Can this approach be applied to different powers, such as \((ax - k)^3(bx - k)^3\)?

**Tip:** Always apply the chain rule when differentiating composite functions like \((ax - k)^2\) and simplify expressions progressively to avoid mistakes.

Explore how to differentiate the expression (ax-k)^2(bx-k)^2 using the product and chain rules. This solution includes detailed steps and explanations, perfect for students in Grades 11-12 looking to strengthen their calculus skills.

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