Differentiation of Functions

Calculus
Name:
Institution:
15th March 2019

Question 1:
(a) State two formulas for the derivative
Solution
 The first formula is the power Rule of Derivatives. According to this formula we
have;
d
dx [ xn ]=n xn−1
 The second formula of the derivative is by the limit. According to this formula we
have;
f ' ( x ) = d
dx [ f (x) ] lim
h → 0
f ( x +h ) −f (x )
h
(b) For f ( x )= 3 x
2−x find f ' ( x ) using the limit definition of a derivative
Solution
The limit definition of the derivative states that for a function f(x) its derivative equals
f ' ( x )=lim
h →0
f ( x+h )−f (x)
h
f ' ( x ) =lim
h →0
1
h ( 3 (x+h)
2−x +h − 3 x
2−x )
¿ lim
h→ 0
1
h ( ( 3 x+3 h ) ( 2−x ) −3 x(2−x+ h)
( 2−x+h)(2−x) )
¿ lim
h→ 0
1
h ( ( 6 x +6 h−3 x2−3 xh ) −(6 x −3 x2 +3 xh)
(2−x+h)(2−x ) )
¿ lim
h→ 0
1
h ( 6 h−6 xh
(2−x+h)(2−x) )
¿ lim
h→ 0
6 ( 1−x
(2−x+ h)(2−x) )
¿ 6
(x−2)( x−2)
¿ 6
( x−2)2

Question 2:
Differentiate the following:
(a) y=7 t 3 +15 et
Solution
y' = dy
dx = dy
dx [ 7 t3 +15 et ]
¿ 7. dy
dt [ t3 ] +15. dy
dt [ et ] =7∗3 t2 +15 et
¿ 21 t2 +15 et
(b) f ( x ) = ( ln ( 3 ) ) x
Solution
f ' ( x )= d
dx [ ( ln (3 ) )x
]
¿ d
dx ¿
¿ e¿¿ ¿
¿∈¿
¿ 1∈¿
¿∈¿
(c) g ( x ) =e9 + xe
Solution
g' ( x)= d
dx [ e9 + xe ]
¿ d
dx [ e9 ]+ d
dx [ xe ]
¿ 0+ d
dx [ xe ]
¿ d
dx [ xe ]
¿ e xe−1

(d) h ( x ) =esin ( 5 x ) −4 √ x
Solution
h' ( x )= d
dx [ esin (5 x )−4 √ x ]
¿ d
dx [ esin ( 5 x ) ]−4 . d
dx [ √ x ]
¿ esin ( 5 x ) d
dx [ sin (5 x ) ]−4 . 1
2 [ x
1
2 −1
]
¿ cos (5 x) d
dx [5 x ] . esin ( 5 x )− 2
√ x
¿ 5 d
dx [ x ] .cos (5 x)esin ( 5 x )− 2
√ x
¿ 5 cos(5 x )esin ( 5 x ) − 2
√ x
(e) d
dx [ ( 2 x5+ 2 ln x )
15
]
Solution
¿ d
dx [ ( 2 x5+2 ln x )15
]
¿ 15 ( 2 x5 +2 ln x ) 14
. d
dx [ ( 2 x5+ 2 ln x ) ]
¿ 15 ( 2 . d
dx [ ln x ] +2 . d
dx [ 2 x5 ] ) ( 2 x5 +2 ln x )
14
¿ 15 ( 2 . 1
x + 2. 5 x4
) ( 2 x5+ 2 ln x )
14
¿ 15 (10 x5 + 2
x ) ( 2 x5 +2 ln x )14
¿ ( 150 x5 +30 ) ( 2 x5+ 2 ln x )
14
x
(f) d
dx [ tan ( e6 x ) ]
Solution
¿ d
dx [ tan ( e6 x ) ]