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Miscellaneous Exercise · Q4

Q.Find the derivative of (ax+b)(cx+d)2(ax + b)(cx + d)^2.

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The derivative of a product of two functions is found using the product rule. For (ax+b)(cx+d)2(ax+b)(cx+d)^2, the derivative is (cx+d)[3acx+ad+2bc](cx+d)\big[3acx + ad + 2bc\big].

Why the Product Rule Works Here

When you see a function written as one expression multiplied by another, the product rule is your natural tool. It says: the derivative of u⋅vu \cdot v is u′v+uv′u'v + uv'. You take turns — differentiate one, leave the other alone, then switch.

Here, u=ax+bu = ax + b (a simple linear function) and v=(cx+d)2v = (cx + d)^2 (a quadratic in disguise). The trick is that vv itself needs the chain rule, because it's something squared. So we'll handle that inner derivative carefully.

Step-by-Step Solution

1. Identify the two pieces.

Let u=ax+bu = ax + b and v=(cx+d)2v = (cx + d)^2.

Then the given function is y=u⋅vy = u \cdot v.

2. Differentiate uu.

u′=au' = a — straightforward, since the derivative of axax is aa and bb is constant.

3. Differentiate vv using the chain rule.

Think of vv as (inside)2(\text{inside})^2 where inside =cx+d= cx + d.

The chain rule says: derivative of (inside)2(\text{inside})^2 is 2(inside)⋅(derivative of inside)2(\text{inside}) \cdot (\text{derivative of inside}).

So v′=2(cx+d)⋅c=2c(cx+d)v' = 2(cx + d) \cdot c = 2c(cx + d).

Tip

A quick check: if you expand (cx+d)2=c2x2+2cdx+d2(cx+d)^2 = c^2x^2 + 2cdx + d^2, then differentiate term-by-term to get 2c2x+2cd=2c(cx+d)2c^2x + 2cd = 2c(cx + d). Same result — but the chain rule is faster.

4. Apply the product rule.

y′=u′v+uv′=a⋅(cx+d)2+(ax+b)⋅2c(cx+d)y' = u'v + uv' = a \cdot (cx + d)^2 + (ax + b) \cdot 2c(cx + d).

5. Factor out the common factor (cx+d)(cx + d). …

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