What is the Derivative of Cos x ? The derivative of cos x is -sin x.
d x d ( cos x ) = − s in x
Proof of Derivative of Cos x Way 1 f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) ( cos x ) ′ = h → 0 lim h cos ( x + h ) − cos x cos ( p + q ) = cos p . cos q − s in p . s in q ( cos x ) ′ = h → 0 lim h cos x . cos h − s in x . s in h − cos x
( cos x ) ′ = h → 0 lim h cos x . cos h − cos x − s in x . s in h
( cos x ) ′ = h → 0 lim h cos x . ( cos h − 1 ) − s in x . s in h
( cos x ) ′ = h → 0 lim [ h cos x . ( cos h − 1 ) − h s in x . s in h ]
( cos x ) ′ = h → 0 lim h cos x . ( cos h − 1 ) − h → 0 lim h s in x . s in h
( cos x ) ′ = cos x . h → 0 lim h ( cos h − 1 ) − s in x . h → 0 lim h s in h
t → 0 l i m t s in t = 1 t → 0 l i m t cos t − 1 = 0
( cos x ) ′ = cos x .0 − s in x .1
( cos x ) ′ = 0 − s in x
( cos x ) ′ = − s in x
Way 2 f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) ( cos x ) ′ = h → 0 lim h cos ( x + h ) − cos x cos p − cos q = − 2. s in ( 2 p − q ) . s in ( 2 p + q ) ( cos x ) ′ = h → 0 lim h − 2. s in ( 2 x + h − x ) . s in ( 2 x + h + x ) ( cos x ) ′ = h → 0 lim h − 2. s in ( 2 h ) . s in ( 2 2 x + h ) ( cos x ) ′ = h → 0 lim h − 2. s in ( 2 h ) . s in [ 2 2 . ( x + 2 h ) ] ( cos x ) ′ = h → 0 lim 2 h − s in ( 2 h ) . s in ( x + 2 h ) ( cos x ) ′ = − h → 0 lim 2 h s in ( 2 h ) . s in ( x + 2 h ) ( cos x ) ′ = − h → 0 lim [ 2 h s in ( 2 h ) . s in ( x + 2 h )] ( cos x ) ′ = − h → 0 lim 2 h s in ( 2 h ) . h → 0 lim s in ( x + 2 h )
h = 2 h ( h → 0 )
( cos x ) ′ = − h → 0 lim h s in h . h → 0 lim s in ( x + h )
t → 0 l i m t s in t = 1
( cos x ) ′ = − 1. s in ( x + 0 )
( cos x ) ′ = − 1. s in x
( cos x ) ′ = − s in x
Way 3
By using the infinite series expansions of the sin x and cos x functions, we can prove that the derivative of cos x is equal to -sin x. The infinite series expansions of the sin x and cos x functions are as follows.
s in x = x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + 9 ! x 9 − ...
cos x = 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + 8 ! x 8 − ...
cos x = 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + 8 ! x 8 − ...
( cos x ) ′ = ( 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + 8 ! x 8 − ... ) ′
( cos x ) ′ = ( 1 ) ′ − ( 2 ! x 2 ) ′ + ( 4 ! x 4 ) ′ − ( 6 ! x 6 ) ′ + ( 8 ! x 8 ) ′ − ...
( cos x ) ′ = 0 − 2 ! 2. x + 4 ! 4. x 3 − 6 ! 6. x 5 + 8 ! 8. x 7 − ...
( cos x ) ′ = − 2 .1 ! 2 . x + 4 .3 ! 4 . x 3 − 6 .5 ! 6 . x 5 + 8 .7 ! 8 . x 7 − ...
( cos x ) ′ = − 1 x + 3 ! x 3 − 5 ! x 5 + 7 ! x 7 − ...
( cos x ) ′ = − x + 3 ! x 3 − 5 ! x 5 + 7 ! x 7 − ...
( cos x ) ′ = − ( x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + ... )
( cos x ) ′ = − s in x
⇓
f ( x ) = cos u ( x ) ⇒ f ′ ( x ) = − u ′ ( x ) . s in u ( x ) ( click to see the proof of the formula )
Q u es t i o n :
f ( x ) = cos ( 5 x 2 + 7 x − 3 ) ⇒ f ′ ( x ) = ?
A n s w er :
f ( x ) = cos ( 5 x 2 + 7 x − 3 )
f ′ ( x ) = [ cos ( 5 x 2 + 7 x − 3 ) ] ′
f ( x ) = cos u ( x ) ⇒ f ′ ( x ) = − u ′ ( x ) . s in u ( x )
f ′ ( x ) = − ( 5 x 2 + 7 x − 3 ) ′ . s in ( 5 x 2 + 7 x − 3 )
f ′ ( x ) = − ( 2.5. x + 7 − 0 ) . s in ( 5 x 2 + 7 x − 3 )
f ′ ( x ) = − ( 10 x + 7 ) . s in ( 5 x 2 + 7 x − 3 )