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dy<br />

dx = dy du<br />

du dx = − 1 du<br />

u√u − 1 dx (terbukti)<br />

Contoh 5.21<br />

Jika y = arccsc x − p 2<br />

Penyelesaian<br />

Misal u = x − p 2<br />

du<br />

dx = 1<br />

dy<br />

dx = dy du<br />

du<br />

, tentukan<br />

dy<br />

dx<br />

y = 2arccot u<br />

dy<br />

du = − 1<br />

u√u − 1<br />

dx = − 1<br />

u√u − 1 (1) = − 1<br />

u√u − 1 = − 1<br />

x − p 2<br />

x − p 2 − 1<br />

Soal-soal<br />

Carilah turunan pertama dari soal-soal berikut!<br />

1. y = arcsin(p − x) 3. y = cos2x<br />

arccos x<br />

2. y = −3arccos 4x 4. y = arctan x − sin3x<br />

5.8 Turunan fungsi eksponen<br />

Jika y = f(x) = e , maka dy = f'(x) = e (5.37)<br />

dx<br />

Bukti<br />

e dide inisikan sebagai lim ®<br />

1 + x n<br />

Dengan menggunakan teorema binomial didapat,<br />

1 + x n<br />

= 1 0!<br />

x<br />

n + (n). 1<br />

1!<br />

x<br />

n<br />

+<br />

n(n − 1). 1<br />

2!<br />

x<br />

n +<br />

n(n − 1)(n − 2). 1<br />

3!<br />

x<br />

n<br />

+ ⋯<br />

= 1 + x + 1 − 1 n<br />

2!<br />

. x + (1 − 1 n )(1 − 2 n )<br />

. x + ⋯<br />

3!<br />

lim ®<br />

1 + x n<br />

= lim ®<br />

1 + x + 1 − 1 n<br />

2!<br />

. x + (1 − 1 n )(1 − 2 n )<br />

. x + ⋯<br />

3!<br />

e = 1 + x + x 2! + x 3! + ⋯ (5.38)<br />

112

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