Sin x'in Türevi Nedir ? Sin x'in türevi cos x'tir.
d x d ( s in x ) = cos x
Sin x'in Türevinin İspatı 1. Yol f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) ( s in x ) ′ = h → 0 lim h s in ( x + h ) − s in x s in ( p + q ) = s in p . cos q + s in q . cos p ( s in x ) ′ = h → 0 lim h s in x . cos h + s in h . cos x − s in x ( s in x ) ′ = h → 0 lim h s in x . cos h − s in x + s in h . cos x ( s in x ) ′ = h → 0 lim h s in x . ( cos h − 1 ) + s in h . cos x ( s in x ) ′ = h → 0 lim [ h s in x . ( cos h − 1 ) + h s in h . cos x ] ( s in x ) ′ = h → 0 lim h s in x . ( cos h − 1 ) + h → 0 lim h s in h . cos x ( s in x ) ′ = s in x . h → 0 lim h ( cos h − 1 ) + cos 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
( s in x ) ′ = s in x .0 + cos x .1
( s in x ) ′ = 0 + cos x
( s in x ) ′ = cos x
2. Yol f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) ( s in x ) ′ = h → 0 lim h s in ( x + h ) − s in x s in p − s in q = 2. s in ( 2 p − q ) . cos ( 2 p + q ) ( s in x ) ′ = h → 0 lim h 2. s in ( 2 x + h − x ) . cos ( 2 x + h + x ) ( s in x ) ′ = h → 0 lim h 2. s in ( 2 h ) . cos ( 2 2 x + h )
( s in x ) ′ = h → 0 lim h 2. s in ( 2 h ) . cos [ 2 2 . ( x + 2 h ) ]
( s in x ) ′ = h → 0 lim 2 h s in ( 2 h ) . cos ( x + 2 h )
( s in x ) ′ = h → 0 lim [ 2 h s in ( 2 h ) . cos ( x + 2 h )]
( s in x ) ′ = h → 0 lim 2 h s in ( 2 h ) . h → 0 lim cos ( x + 2 h )
h = 2 h ( h → 0 )
( s in x ) ′ = h → 0 lim h s in h . h → 0 lim cos ( x + h )
t → 0 l i m t s in t = 1
( s in x ) ′ = 1. cos ( x + 0 )
( s in x ) ′ = 1. cos x
( s in x ) ′ = cos x
3. Yol Sin x ve cos x fonksiyonlarının sonsuz seri şeklindeki açılımlarından faydalanarak da sin x'in türevinin cos x'e eşit olduğunu ispatlayabiliriz. Sin x ve cos x fonksiyonlarının sonsuz seri şeklindeki açılımları aşağıdaki gibidir.
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 − ...
s in x = x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + 9 ! x 9 − ...
( s in x ) ′ = ( x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + 9 ! x 9 − ... ) ′
( s in x ) ′ = ( x ) ′ − ( 3 ! x 3 ) ′ + ( 5 ! x 5 ) ′ − ( 7 ! x 7 ) ′ + ( 9 ! x 9 ) ′ − ...
( s in x ) ′ = 1 − 3 ! 3. x 2 + 5 ! 5. x 4 − 7 ! 7. x 6 + 9 ! 9. x 8 − ...
( s in x ) ′ = 1 − 3 .2 ! 3 . x 2 + 5 .4 ! 5 . x 4 − 7 .6 ! 7 . x 6 + 9 .8 ! 9 . x 8 − ...
( s in x ) ′ = 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + 8 ! x 8 − ...
( s in x ) ′ = cos x
⇓
f ( x ) = s in u ( x ) ⇒ f ′ ( x ) = u ′ ( x ) . cos u ( x ) ( formülün ispatını görmek için tıklayınız )
S or u :
f ( x ) = s in ( x 3 − 5 x 2 + 2 x + 7 ) ⇒ f ′ ( x ) = ?
C e v a p :
f ( x ) = s in ( x 3 − 5 x 2 + 2 x + 7 )
f ′ ( x ) = [ s in ( x 3 − 5 x 2 + 2 x + 7 ) ] ′
f ( x ) = s in u ( x ) ⇒ f ′ ( x ) = u ′ ( x ) . cos u ( x )
f ′ ( x ) = ( x 3 − 5 x 2 + 2 x + 7 ) ′ . cos ( x 3 − 5 x 2 + 2 x + 7 )
f ′ ( x ) = ( 3. x 2 − 5.2. x + 2 + 0 ) . cos ( x 3 − 5 x 2 + 2 x + 7 )
f ′ ( x ) = ( 3 x 2 − 10 x + 2 ) . cos ( x 3 − 5 x 2 + 2 x + 7 )
S or u :
f ( x ) = s in 2 x + s in x 3 ⇒ f ′ ( 0 ) = ?
C e v a p :
f ( x ) = s in 2 x + s in x 3
f ′ ( x ) = ( s in 2 x + s in x 3 ) ′
f ′ ( x ) = ( s in 2 x ) ′ + ( s in x 3 ) ′
f ( x ) = s in u ( x ) ⇒ f ′ ( x ) = u ′ ( x ) . cos u ( x )
f ′ ( x ) = ( 2 x ) ′ . cos 2 x + ( x 3 ) ′ . cos x 3
f ′ ( x ) = 2. cos 2 x + 3. x 2 . cos x 3
f ′ ( 0 ) = 2. cos ( 2.0 ) + 3. 0 2 . cos ( 0 3 )
f ′ ( 0 ) = 2. cos 0 + 3.0. cos 0
f ′ ( 0 ) = 2. cos 0 + 0
f ′ ( 0 ) = 2. cos 0
cos 0 = 1
f ′ ( 0 ) = 2.1
f ′ ( 0 ) = 2
S or u :
f ( x ) = s i n 2 x ⇒ f ( x ) f ′ ( x ) = ?
C e v a p :
f ( x ) = s i n 2 x
f ′ ( x ) = ( s i n 2 x ) ′
f ( x ) = u 2 ⇒ f ′ ( x ) = 2. u . u ′
f ′ ( x ) = 2. s in x . ( s in x ) ′
f ′ ( x ) = 2. s in x . cos x
f ( x ) f ′ ( x ) = s i n 2 x 2. s in x . cos x
f ( x ) f ′ ( x ) = s in x . s in x 2. s in x . cos x
f ( x ) f ′ ( x ) = 2. s in x cos x
s in x cos x = co t x
f ( x ) f ′ ( x ) = 2 co t x