[u(x).v(x)]'=u'(x).v(x)+v'(x).u(x)'tir.
İki Fonksiyonun Çarpımının Türevi Nedir ? [ u ( x ) . v ( x ) ] ′ = u ′ ( x ) . v ( x ) + v ′ ( x ) . u ( x )
d x d [ u ( x ) . v ( x )] = d x d [ u ( x )] . v ( x ) + d x d [ v ( x )] . u ( x )
Çarpımın Türevinin İspatı 1. Yol f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x )
[ u ( x ) . v ( x ) ] ′ = h → 0 lim h u ( x + h ) . v ( x + h ) − u ( x ) . v ( x )
[ u ( x ) . v ( x ) ] ′ = h → 0 lim h u ( x + h ) . v ( x + h ) − u ( x ) . v ( x ) + u ( x ) . v ( x + h ) − u ( x ) . v ( x + h )
[ u ( x ) . v ( x ) ] ′ = h → 0 lim h u ( x + h ) . v ( x + h ) − u ( x ) . v ( x + h ) + u ( x ) . v ( x + h ) − u ( x ) . v ( x )
[ u ( x ) . v ( x ) ] ′ = h → 0 lim h v ( x + h ) . [ u ( x + h ) − u ( x )] + u ( x ) . [ v ( x + h ) − v ( x )]
[ u ( x ) . v ( x ) ] ′ = h → 0 lim { h v ( x + h ) . [ u ( x + h ) − u ( x )] + h u ( x ) . [ v ( x + h ) − v ( x )] }
[ u ( x ) . v ( x ) ] ′ = h → 0 lim h v ( x + h ) . [ u ( x + h ) − u ( x )] + h → 0 lim h u ( x ) . [ v ( x + h ) − v ( x )]
[ u ( x ) . v ( x ) ] ′ = h → 0 lim [ v ( x + h ) . h u ( x + h ) − u ( x ) ] + h → 0 lim [ u ( x ) . h v ( x + h ) − v ( x ) ]
[ u ( x ) . v ( x ) ] ′ = h → 0 lim v ( x + h ) . h → 0 lim h u ( x + h ) − u ( x ) + h → 0 lim u ( x ) . h → 0 lim h v ( x + h ) − v ( x )
[ u ( x ) . v ( x ) ] ′ = v ( x + 0 ) . u ′ ( x ) + u ( x ) . v ′ ( x )
[ u ( x ) . v ( x ) ] ′ = v ( x ) . u ′ ( x ) + u ( x ) . v ′ ( x )
[ u ( x ) . v ( x ) ] ′ = u ′ ( x ) . v ( x ) + v ′ ( x ) . u ( x )
2. Yol f ( x ) = u ( x ) . v ( x )
l n f ( x ) = l n [ u ( x ) . v ( x )]
l n f ( x ) = l n u ( x ) + l n v ( x )
( l n u ) ′ = u u ′
f ( x ) f ′ ( x ) = u ( x ) u ′ ( x ) + v ( x ) v ′ ( x )
f ( x ) f ′ ( x ) = u ( x ) . v ( x ) u ′ ( x ) . v ( x ) + v ′ ( x ) . u ( x )
f ′ ( x ) = u ′ ( x ) . v ( x ) + v ′ ( x ) . u ( x )
3. Yol y = f ( x )
y + d y = f ( x + d x )
y + d y − y = f ( x + d x ) − f ( x )
d y = f ( x + d x ) − f ( x )
d x d y = d x f ( x + d x ) − f ( x )
d x d y = d x → 0 lim d x f ( x + d x ) − f ( x )
d x d y = f ′ ( x )
d x d y = y ′
y = u . v
y + d y = ( u + d u ) . ( v + d v )
y + d y = u . v + u . d v + v . d u + d u . d v
y + d y − y = u . v + u . d v + v . d u + d u . d v − u . v
d y = u . d v + v . d u + d u . d v
d y = u . d v + v . d u + ihmal edilir d u . d v
d y = u . d v + v . d u
d x d y = d x u . d v + d x v . d u
d x d y = u . d x d v + v . d x d u
y ′ = u . v ′ + v . u ′
y ′ = u ′ . v + v ′ . u
4. Yol ( u + v ) 2 = u 2 + 2. u . v + v 2
( u + v ) 2 − u 2 − v 2 = 2. u . v
2 ( u + v ) 2 − u 2 − v 2 = 2 2 . u . v
u . v = 2 ( u + v ) 2 − u 2 − v 2
( u . v ) ′ = [ 2 ( u + v ) 2 − u 2 − v 2 ] ′
( u . v ) ′ = 2 [( u + v ) 2 ] ′ − ( u 2 ) ′ − ( v 2 ) ′
( u 2 ) ′ = 2. u . u ′
( u . v ) ′ = 2 2. ( u + v ) . ( u + v ) ′ − 2. u . u ′ − 2. v . v ′
( u . v ) ′ = 2 2. ( u + v ) . ( u ′ + v ′ ) − 2. u . u ′ − 2. v . v ′
( u . v ) ′ = 2 2. ( u . u ′ + u . v ′ + v . u ′ + v . v ′ ) − 2. u . u ′ − 2. v . v ′
( u . v ) ′ = 2 2. u . u ′ + 2. u . v ′ + 2. v . u ′ + 2. v . v ′ − 2. u . u ′ − 2. v . v ′
( u . v ) ′ = 2 2. u . v ′ + 2. v . u ′
( u . v ) ′ = 2 2 . ( u . v ′ + v . u ′ )
( u . v ) ′ = u . v ′ + v . u ′
( u . v ) ′ = u ′ . v + v ′ . u
Soru f ( x ) = ( 3 x − 4 ) . ( 5 x + 7 ) ⇒ f ′ ( x ) = ?
Cevap f ′ ( x ) = [( 3 x − 4 ) . ( 5 x + 7 ) ] ′
f ′ ( x ) = ( 3 x − 4 ) ′ . ( 5 x + 7 ) + ( 5 x + 7 ) ′ . ( 3 x − 4 )
f ′ ( x ) = 3. ( 5 x + 7 ) + 5. ( 3 x − 4 )
f ′ ( x ) = 3.5 x + 3.7 + 5.3 x − 5.4
f ′ ( x ) = 15 x + 21 + 15 x − 20
f ′ ( x ) = 15 x + 15 x + 21 − 20
f ′ ( x ) = 30 x + 1