Page 271 - Matematika_XI_Siswa

c. Turunan fungsi f(x) = [u(x)] dengan u'(x) ada, n bilangan asli.
n
()
) x
f '(x) = lim ( fx +D - fx
D→ 0 D x
x
n
ux
= lim [(ux +D x )] - [()] n
D→ 0 D x
x
n
ux
() ux
) x
= lim [(ux +D - ux + ()] - [()] n
x
D→ 0
Misal P = [u(x + Dx) – u(x)]
ux
()] -
+
= lim [P ux n [()] n (Gunakan Binomial Newton)
D→ 0 D x
x
ux
ux
n
ux
+
2
n
ux
n
... C P
n
n
ux
= lim P + C P n - 1 [()] C P n - 2 [()] ++ n- 1 [()] n - 1 + [()] - [()] n
2
1
D→ 0 D x
x
... C
+
n
ux
2
n
ux
ux
n
ux
= lim P + nP n - 1 [()] C P n - 2 [()] ++ n n - 2 P 2 [()] n - 2 + C P [()] n - 1
1
n
-
2
x
D→ 0 D x
-
1
n
ux
ux
2
ux
... C
= lim ( P P n - 1 + nP n - 2 [()] ++ n n - 2 P [()] n - 2 + C n n - 1 [()] )
D→ 0 D x
x
1
2
= lim P lim(P n - 1 + nP n - 2 [()] ++ n P [()] n - 2 + C n [()] )
-
n
ux
... C
ux
ux
D→ 0 D x D→ 0 n- 2 n- 1
x
x
) x
()
Karena lim P = lim ( ux +D - ux = u'(x)
D→ 0 D x D→ 0 D x
x
x
lim P = lim u(x + Dx) – u(x) = 0
D→ 0 D→ 0
x
x
= u'(x)[0 + n[u(x)]] n – 1
= nu'(x)[u(x)] n – 1 .
MATEMATIKA 261