Question

12. (ax +b)"

Answer 1

Consider the function,

f( x )= ( ax+b ) n

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

f ′ ( x )= lim h→0 f( x+h )−f( x ) h

Apply the above formula in the given function,

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

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

Thus, the derivative of ( ax+b ) n is na ( ax+b ) n−1 .