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 ) :-

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

Answer it plz ! ​

Answer 1

$\large \bigstar \underline{ \pmb{ \frak{Required \: Solution : - }}}$

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

hope its helps !

Answer 2

Given :-

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

To Find :-

The derivative of ${ ( ax + b ) }^{n} { ( cx + d ) }^{m}$ satisfying the given condition .

Used Concepts :-

• Leibnitz's Product rule of differentiation which states that for any two functions ( u and v ) ; $\frac{d}{dx} ( u . v ) = u . \frac{dv}{dx} + v . \frac{du}{dx}$
• $\frac{d}{dx} x^{n} = n.{x}^{n - 1 }$
• Chain Rule of differentiation

Solution :-

Let us assume that ;

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

Py Product Rule Differentiating both sides w.r.t."x" ;

=> $\frac{d}{dx} ( y ) = \frac{d}{dx} [ { ( ax + b ) }^{n} . { ( cx + d ) }^{m} ]$

=> $\frac{dy}{dx} = { ( ax + b ) }^{n} . \frac{d}{dx} ( cx + d )^{m} + { ( cx + d ) }^m . \frac{d}{dx} ( ax + b )^{n}$

=> Applying Chain Rule and the second concept above we get ;

=> $\frac{dy}{dx} = { ( ax + b ) }^{n} . m{ ( cx + d ) }^{m - 1} . \frac{d}{dx} ( cx + d ) + { ( cx + d ) }^{m} . n{ ( ax + b ) }^{n - 1} .\frac{d}{dx}( ax + b )$

=> $\frac{dy}{dx} = { ( ax + b ) }^{n} . m{ ( cx + d ) }^{m-1} . [ \frac{d}{dx} ( cx ) + \frac{d}{dx} ( d ) ] + { ( cx + d ) }^{m} . n{ ( ax + b ) }^{n-1} . [ \frac{d}{dx} ( ax ) + \frac{d}{dx} ( b )$

=> As a , b , c and d are constant . So ;

=> $\frac{dy}{dx} = { ( ax + b ) }^{n} . m{ ( cx + d ) }^{m-1} . [ c . \frac{d}{dx} ( x ) + 0 ] + { ( cx + d ) }^{m} . n{ ( ax + b ) }^{n-1} . [ a . \frac{d}{dx} ( x ) + 0 ]$

=> $\frac{dy}{dx} = { ( ax + b ) }^{n} . m{ ( cx + d ) }^{m-1} . c + { ( cx + d ) }^{m} . n{ ( ax + b ) }^{n-1} . a$

=> $\frac{dy}{dx} = { ( ax + b ) }^{n} . cm{ ( cx + d ) }^{m-1} + { ( cx + d ) }^{m} . an{ ( ax + b ) }^{n-1}$

Now. , We are Done :)