(i)
Consider the given function,
f( x )=( 2x− 3 4 )
We know that the derivative of a function f( x ) is f ′ ( x )and represented as,
f ′ ( x )= d dx f( x )
The formula for the derivative of the function is,
d( x n ) dx =( n x n−1 )
Use the above formula to find the derivative of given function.
d dx ( 2x− 3 4 )= d dx ( 2x )− d dx ( 3 4 ) =2−0 =2
Thus, the derivative of ( 2x− 3 4 )is 2.
(ii)
Consider the given function,
f( x )=( 5 x 3 +3x−1 )( x−1 )
We know that the derivative of a function f( x ) is f ′ ( x )and represented as,
f ′ ( x )= d dx f( x )
The Leibnitz product rule to find the derivative of the function is,
d dx ( U⋅V )=V⋅ U ′ +U⋅ V ′
Where U ′ and V ′ are the derivatives of their respective functions.
Use the above formula to find the derivative of given function.
d dx [ ( 5 x 3 +3x−1 )⋅( x−1 ) ]=( x−1 ) d dx ( 5 x 3 +3x−1 )+( 5 x 3 +3x−1 ) d dx ( x−1 ) =( x−1 )( 5⋅3 x 2 +3−0 )+( 5 x 3 +3x−1 )( 1−0 ) =( x−1 )( 15 x 2 +3 )+1( 5 x 3 +3x−1 ) =( 15 x 3 +3x−15 x 2 −3 )+5 x 3 +3x−1
Further simplify,
d dx [ ( 5 x 3 +3x−1 )⋅( x−1 ) ]=( 20 x 3 −15 x 2 +6x−4 )
Thus, the derivative of ( 5 x 3 +3x−1 )( x−1 ) is ( 20 x 3 −15 x 2 +6x−4 ).
(iii)
Consider the given function,
f( x )= x −3 ( 5+3x )
We know that the derivative of a function f( x ) is f ′ ( x )and represented as,
f ′ ( x )= d dx f( x )
The Leibnitz product rule to find the derivative of the function is,
d dx ( U⋅V )=V⋅ U ′ +U⋅ V ′
Where U ′ and V ′ are the derivatives of their respective functions
Use the above formula to find the derivative of given function.
d dx [ ( x −3 )( 5+3x ) ]=( 5+3x ) d dx ( x −3 )+ x −3 d dx ( 5+3x ) =( 5+3x )( −3 x −4 )+ x −3 ( 0+3 ) =−15 x −4 −9 x −3 +3 x −3 =−15 x −4 −6 x −3
Simplify further,
d dx [ ( x −3 )( 5+3x ) ]=−3 x −3 ( 2+5 x −1 ) =− 3 x 3 ( 2+ 5 x ) =− 3 x 3 ( 2x+5 x ) =− 3 x 4 ( 2x+5 )
Thus, the derivative of x −3 ( 5+3x ) is − 3 x 4 ( 2x+5 ).
(iv)
Consider the given function,
f( x )= x 5 ( 3−6 x −9 )
We know that the derivative of a function f( x ) is f ′ ( x )and represented as,
f ′ ( x )= d dx f( x )
The Leibnitz product rule to find the derivative of the function is,
d dx ( U⋅V )=V⋅ U ′ +U⋅ V ′
Where U ′ and V ′ are the derivatives of their respective functions.
Use the above formula to find the derivative of given function.
d dx x 5 ( 3−6 x −9 )= x 5 d dx ( 3−6 x −9 )+( 3−6 x −9 ) d dx x 5 = x 5 ( 0−6( −9 x −10 ) )+( 3−6 x −9 )5 x 4 = x 5 ( 54 x −10 )+15 x 4 −30 x −5 =54 x −5 +15 x 4 −30 x −5
Simplify further,
d dx x 5 ( 3−6 x −9 )=15 x 4 +24 x −5 =15 x 4 + 24 x 5
Thus, the derivative of x 5 ( 3−6 x −9 ) is 15 x 4 + 24 x 5 .
(v)
Consider the given function,
f( x )= x −4 ( 3−4 x −5 )
We know that the derivative of a function f( x ) is f ′ ( x )and represented as,
f ′ ( x )= d dx f( x )
The Leibnitz product rule to find the derivative of the function is,
d dx ( U⋅V )=V⋅ U ′ +U⋅ V ′
Where U ′ and V ′ are the derivatives of their respective functions.
Use the above formula to find the derivative of given function.
d dx [ ( x ) −4 ( 3−4 x −5 ) ]=( 3−4 x −5 ) d dx ( x −4 )+ x −4 d dx ( 3−4 x −5 ) =( 3−4 x −5 )( −4 x −5 )+ x −4 ( 0−( −4⋅5 x −6 ) ) =−12 x −5 +16 x −10 +20 x −10 =−12 x −5 +36 x −10
Further simplify,
d dx [ ( x ) −4 ( 3−4 x −5 ) ]= 36 x 10 − 12 x 5
Thus, the derivative of x −4 ( 3−4 x −5 ) is ( 36 x 10 − 12 x 5 ).
(vi)
Consider the given function,
f( x )= 2 x+1 − x 2 3x−1
The quotient rule of derivative to find the derivative of the function is,
d dx ( U V )= ( U V ′ −V U ′ ) V 2
Where U ′ and V ′ are the derivative of their respective functions.
Use the above formula to find the derivative of given function.
d dx ( 2 x+1 − x 2 3x−1 )= d dx ( 2 x+1 )− d dx ( x 2 3x−1 ) =[ ( x+1 ) d dx ( 2 )−2 d dx ( x+1 ) ( x+1 ) 2 − ( 3x−1 ) d dx ( x 2 )− x 2 d dx ( 3x−1 ) ( 3x−1 ) 2 ] =[ ( x+1 )0−2( 1 ) ( x+1 ) 2 − ( 3x−1 )( 2x )− x 2 ( 3 ) ( 3x−1 ) 2 ] =[ 0−2 ( x+1 ) 2 − ( 6 x 2 −2x−3 x 2 ) ( 3x−1 ) 2 ]
Simplify further,
d dx ( 2 x+1 − x 2 3x−1 )=[ −2 ( x+1 ) 2 − ( 3 x 2 −2x ) ( 3x−1 ) 2 ] = −2 ( x+1 ) 2 −x ( 3x−2 ) ( 3x−1 ) 2
Thus, the derivative of 2 x+1 − x 2 3x−1 is −2 ( x+1 ) 2 −x ( 3x−2 ) ( 3x−1 ) 2 .