Question

Q1.Solve pls

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Answer 1

Step-by-step explanation:

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 :

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

By applying the product rule of differentiation, we get :

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

$</p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = [(cx + d)^{m}\cdot\dfrac{d[(ax + b)^{n}]}{dx} + [(ax + b)^{n}\cdot\dfrac{d[(cx + d)^{m}]}{dx}} \\ \\ \\ \end{gathered}$

First,

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

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(ax + b)^{n}]}{dx}} \\ \\ \end{gathered} </p><p>$

By applying the chain rule of differentiation, we get :

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

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(ax + b)^{n}]}{d(ax + b)}\cdot\dfrac{d(ax + b)}{dx}} \\ \\ \\ \end{gathered} </p><p></p><p>$

$</p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot\dfrac{d(ax + b)}{dx}} \\ \\ \\ \end{gathered}$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot\dfrac{d(ax)}{dx} + \dfrac{d(b)}{dx}} \\ \\ \\ \end{gathered} </p><p></p><p>$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot(a + 0)} \\ \\ \\ \end{gathered} </strong></p><p></p><p><strong>[tex]\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = n(ax + b)^{(n - 1)}\cdot(a + 0)} \\ \\ \\ \end{gathered}$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = an(ax + b)^{(n - 1)}} \\ \\ \end{gathered} </p><p>$

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

Now, the derivative of (cx + d)^m :

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(cx + d)^{m}]}{dx}} \\ \\ \end{gathered} </p><p>$

By applying the chain rule of differentiation, we get :

$\begin{gathered}\underline{\sf{\bigstar\: Chain\:Rule\:of\: Differentiation :}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}} \\ \\ \\ \end{gathered} </p><p></p><p>$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(cx + d)^{m}]}{d(cx + d)}\cdot\dfrac{d(cx + d)}{dx}} \\ \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = m(cx + d)^{(m - 1)}\cdot\dfrac{d(cx + d)}{dx}} \\ \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = m(cx + d)^{(m - 1)}\cdot\dfrac{d(cx)}{dx} + \dfrac{d(d)}{dx}} \\ \\ \\ \end{gathered} </p><p></p><p> </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = m(cx + d)^{(m - 1)}\cdot(c + 0)} \\ \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = cm(cx + d)^{(m - 1)}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}\boxed{\therefore \sf{\dfrac{d[(cx + d)^{m}]}{dx} = cm(cx + d)^{(m - 1)}}} \\ \\ \end{gathered} </p><p></p><p>$

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

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = [(cx + d)^{m}\cdot\dfrac{d[(ax + b)^{n}]}{dx} + [(ax + b)^{n}\cdot\dfrac{d[(cx + d)^{m}]}{dx}} \\ \\ \\ \end{gathered} </p><p>$

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

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

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

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

Answer 2

Given :

a, b, c and d are fixed non-zero constants and m, n are integers.

To Find :

The derivative of (ax + b)" (cx + d)m satisfying the given condition.

SOLUTION:-

Given that :

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

where, a and b are constants; m and n are integers.

we are asked to find its derivative.

$\\ \maltese \underline{ \sf \: Product \: rule : - }$

$\\ : \implies \sf y = f(x) \times g(x)$

$\\ \sf : \implies\frac{dy}{dx} = f(x) \times \frac{d}{dx}[g(x) ] + g(x) \times \frac{d}{dx} [f(x) ]$

Applying the same concept :-

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

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

Now, let's find out the derivative of (cx+d)^m

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

$\\ : \implies \boxed{\sf \dfrac{d}{dx} [ {(cx + d)}^{m} ] = cm {(cx + d)}^{(m - 1)}} - - - (2)$

Now, let's find out the derivative of (ax+b)^n

$\\ \\ : \implies \sf \dfrac{d}{dx} [ {(ax + b)}^{n} ] = n. {(ax + d)}^{(n - 1)} .(a + 0)$

$\\ : \implies \boxed{\sf \dfrac{d}{dx} [ {(ax +b)}^{n} ] = an {(ax + b)}^{(n - 1)}} - - - (3)$

Substituting the values of eq. 2 and eq. 3 in eq. 1

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

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

That's the required derivative