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Miscellaneous Exercise · Q13

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

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The derivative is found by applying the product rule and the chain rule to each factor. The final result is (ax+b)n−1(cx+d)m−1[(an)(cx+d)+(cm)(ax+b)](ax+b)^{n-1}(cx+d)^{m-1}\big[(an)(cx+d)+(cm)(ax+b)\big].

The problem asks for the derivative of a product of two functions, each raised to a power. When you see a product of powers of linear expressions, your first instinct should be: product rule combined with chain rule. There is no shortcut formula here — you must differentiate each factor carefully.

Let’s denote:

f(x)=(ax+b)nandg(x)=(cx+d)mf(x) = (ax + b)^n \quad \text{and} \quad g(x) = (cx + d)^m

We want ddx[f(x)⋅g(x)]\frac{d}{dx}[f(x) \cdot g(x)].


1. Recall the product rule

If h(x)=f(x)⋅g(x)h(x) = f(x) \cdot g(x), then

h′(x)=f′(x)g(x)+f(x)g′(x)h'(x) = f'(x) g(x) + f(x) g'(x)

So we need f′(x)f'(x) and g′(x)g'(x) separately.

2. Differentiate f(x)=(ax+b)nf(x) = (ax + b)^n

This is a composite function: outer function unu^n, inner function u=ax+bu = ax + b. By the chain rule:

f′(x)=n(ax+b)n−1⋅ddx(ax+b)=n(ax+b)n−1⋅af'(x) = n(ax + b)^{n-1} \cdot \frac{d}{dx}(ax + b) = n(ax + b)^{n-1} \cdot a

So f′(x)=an(ax+b)n−1f'(x) = a n (ax + b)^{n-1}.

3. Differentiate g(x)=(cx+d)mg(x) = (cx + d)^m

Exactly the same logic:

g′(x)=m(cx+d)m−1⋅ddx(cx+d)=m(cx+d)m−1⋅cg'(x) = m(cx + d)^{m-1} \cdot \frac{d}{dx}(cx + d) = m(cx + d)^{m-1} \cdot c

So g′(x)=cm(cx+d)m−1g'(x) = c m (cx + d)^{m-1}.

Tip

Notice the pattern: derivative of (linear)k(linear)^k is k⋅(linear)k−1⋅(coefficient of x)k \cdot (linear)^{k-1} \cdot \text{(coefficient of }x). This saves time in any chain rule problem.

4. Apply the product rule

h′(x)=[an(ax+b)n−1]⋅(cx+d)m  +  (ax+b)n⋅[cm(cx+d)m−1]h'(x) = \big[a n (ax + b)^{n-1}\big] \cdot (cx + d)^m \;+\; (ax + b)^n \cdot \big[c m (cx + d)^{m-1}\big]

5. Factor common terms

Both terms contain (ax+b)n−1(ax + b)^{n-1} and (cx+d)m−1(cx + d)^{m-1}. Factor them out:

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