Find the derivative of the following function ( it is to be understood

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

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 ) :-

Answer»

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 :

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

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