Question

13. (ax +b)" (cx +d)

Answer 1

Consider the function,

f( x )= ( ax+b ) n ( cx+d ) m

According to the first principle, the derivative of the function is,

f ′ ( x )= lim h→0 f( x+h )−f( x ) h

The given function can be written as,

f( x )= f 1 ( x ) f 2 ( x )

The Leibnitz product rule of derivatives to find the derivative of the function is,

f ′ ( x )= f 1 ( x ) f ′ 2 ( x )+ f 2 ( x ) f ′ 1 ( x ) (1)

On solving the value of f ′ 2 ( x ), we get,

f ′ 2 ( x )= lim h→0 [ ( c( x+h )+d ) m − ( cx+d ) m ] h = lim h→0 [ ( cx+ch+d ) m − ( cx+d ) m ] h = lim h→0 ( cx+d ) m [ ( 1+ ch cx+d ) m −1 ] h = ( cx+d ) m lim h→0 [ ( 1+ ch ax+d ) m −1 ] h

The binomial expansion for mnumber of terms is,

( 1+x ) m =1+mx+ m( m−1 ) 2! x 2 +…

Use the expansion in the function of derivatives,

f ′ 2 ( x )= ( cx+d ) m lim h→0 1 h [ { 1+m( ch cx+d )+ m( m−1 ) 2! ( ch cx+d ) 2 +… }−1 ]

Expand the function to its highest degree,

f ′ 2 ( x )= ( cx+d ) m lim h→0 1 h [ m( ch cx+d )+ m( m−1 ) c 2 h 2 2! ( cx+d ) 2 +… ] = ( cx+d ) m lim h→0 [ m( c cx+d )+ m( m−1 ) c 2 h 2! ( cx+d ) 2 +… ]

Apply the limits,

f ′ 2 ( x )= ( cx+d ) m [ mc cx+d +0 ] =mc ( cx+d ) m ( cx+d ) =mc ( cx+d ) m−1

Solve the value of f ′ 1 ( x ),

f ′ 1 ( x )= lim h→0 [ ( a( x+h )+b ) n − ( ax+b ) n ] h = lim h→0 [ ( ax+ah+b ) n − ( ax+b ) n ] h = lim h→0 ( ax+b ) n [ ( 1+ ah ax+b ) n −1 ] h

From the binomial expansion for n number of terms is,

( 1+x ) n =1+nx+ n( n−1 ) 2! x 2 +…

Use the expansion in the function of derivative,

f ′ 1 ( x )= ( ax+b ) n lim h→0 1 h [ { 1+n( ah ax+b )+ n( n−1 ) 2! ( ah ax+b ) 2 +… }−1 ]

Expand the function to its highest degree,

f ′ 1 ( x )= ( ax+b ) n lim h→0 1 h [ n( ah ax+b )+ n( n−1 ) a 2 h 2 2! ( ax+b ) 2 +… ] = ( ax+b ) n lim h→0 [ n( a ax+b )+ n( n−1 ) a 2 h 2! ( ax+b ) 2 +… ]

Apply the limits,

f ′ 1 ( x )= ( ax+b ) n [ na ax+b +0 ] =na ( ax+b ) n ( ax+b ) =na ( ax+b ) n−1

Put the values of f ′ 1 ( x ) and f ′ 2 ( x ) in equation (1),

f ′ ( x )= ( ax+b ) n mc ( cx+d ) m−1 + ( cx+d ) m na ( ax+b ) n−1 = ( ax+b ) n−1 ( cx+d ) m−1 [ mc( ax+b )+na( cx+d ) ]

Thus, the derivative of ( ax+b ) n ( cx+d ) m is ( ax+b ) n−1 ( cx+d ) m−1 [ mc( ax+b )+na( cx+d ) ].