Question

Find the derivative of the following function ( it is to be understood that a, b, c and d are fixed non-zero constants and m and n are integers ) :-
$\sf \: (ax + b) ^{n} ( {cx + d})^{n}$
​

Answer 1

Question :

Find the derivative of the following function ( it is to be understood that a, b, c and d are fixed non-zero constants and m and n are integers ) :-

$\sf \: (ax + b) ^{n} ( {cx + d})^{m}$

Solution :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(ax + b)^{n}\cdot(cx + d)^{m}]}{dx}}$

By applying the product rule of differentiation, we get :

$\underline{\sf{\bigstar\:Product\:Rule\:of\: Differentiation :}} \\ \\ :\implies\sf{\dfrac{d(vu)}{dx} = v\cdot\dfrac{du}{dx} + u\cdot\dfrac{dv}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = [(cx + d)^{m}\cdot\dfrac{d[(ax + b)^{n}]}{dx} + [(ax + b)^{n}\cdot\dfrac{d[(cx + d)^{m}]}{dx}}$

First,

• let us find the derivative of (ax + b)^n :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(ax + b)^{n}]}{dx}}$

By applying the chain rule of differentiation, we get :

$\underline{\sf{\bigstar\: Chain\:Rule\:of\: Differentiation :}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(ax + b)^{n}]}{d(ax + b)}\cdot\dfrac{d(ax + b)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot\dfrac{d(ax + b)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot\dfrac{d(ax)}{dx} + \dfrac{d(b)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot(a + 0)}$

$:\implies \sf{\dfrac{dy}{dx} = an(ax + b)^{(n - 1)}}$

$\boxed{\therefore \sf{\dfrac{d[(ax + b)^{n}]}{dx} = an(ax + b)^{(n - 1)}}}$

• Now, the derivative of (cx + d)^m :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(cx + d)^{m}]}{dx}}$

By applying the chain rule of differentiation, we get :

$\underline{\sf{\bigstar\: Chain\:Rule\:of\: Differentiation :}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(cx + d)^{m}]}{d(cx + d)}\cdot\dfrac{d(cx + d)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = m(cx + d)^{(m - 1)}\cdot\dfrac{d(cx + d)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = m(cx + d)^{(m - 1)}\cdot\dfrac{d(cx)}{dx} + \dfrac{d(d)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = m(cx + d)^{(m - 1)}\cdot(c + 0)}$

$:\implies \sf{\dfrac{dy}{dx} = cm(cx + d)^{(m - 1)}}$

$\boxed{\therefore \sf{\dfrac{d[(cx + d)^{m}]}{dx} = cm(cx + d)^{(m - 1)}}}$

Now by substituting the derivative of (ax + b)^n and (cx + d)^m in the equation, we get :

$:\implies \sf{\dfrac{dy}{dx} = [(cx + d)^{m}\cdot\dfrac{d[(ax + b)^{n}]}{dx} + [(ax + b)^{n}\cdot\dfrac{d[(cx + d)^{m}]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = [(cx + d)^{m}\cdot an(ax + b)^{(n - 1)}] + [(ax + b)^{n}\cdot cm(cx + d)^{(m - 1)}]}$

By taking (ax + b)^(n - 1) and (cx + d)^(m - 1) common in the equation, we get :

$:\implies \sf{\dfrac{dy}{dx} = [cm(cx + d)^{m}\cdot an(ax + b)^{n}] + [(ax + b)^{n - 1)}\cdot(cx + d)^{(m - 1)}]}$

$\underline{\therefore \sf{\dfrac{dy}{dx} = [cm(cx + d)^{m}\cdot an(ax + b)^{n}] + [(ax + b)^{n - 1)}\cdot(cx + d)^{(m - 1)}]}}$

Answer 2

⭐Answer⭐

Let f(x) = (ax + b) (cx + d)^2

Thus by leibnitz product rule.

f'(x) = (ax + b) d/dx (cx + d)^2 + (cx + d)^2 d/dx (ax + b).

(ax + b) d/dx (c^2x^2 + 2 cdx + d^2) + (cx+d)^2 d/dx (ax + b)

(ax + b) [d/dx (c^2 x^2) + d/dx (2cdx) + d/dx d^2]

(cx + d)^2 [d/dx ax + d/dx b]

(ax + b) (2c^2 x + 2cd) + (cx + d^2)a

2a(ax + b) (cx + d) + a(cx + d)^2

Hope it helps❤