Derivadas 2

Page 1

1. Determine, si existe, la derivada de f en x = 2 de la funci´ on f definida por  2 ,  x +1 f (x) = 5 ,  (x − 1)2 + 4 ,

Soluci´ on

l´ım−

h→0

=

f (2 + h) − f (2) = h

l´ım−

h→0

h→0−

l´ım

h→0+

=

l´ım

l´ım

h→0+

((2 + h − 1)2 + 4) − (5) h

(1 + 2h + h2 + 4) − (5) h

= l´ım+

h(2 + h) = h

como:

l´ım

h→0

l´ım (4 + h) = 4

h→0−

f (2 + h) − f (2) = h

h→0+

h→0

((2 + h)2 + 1) − (5) h

(22 + 4h + h2 + 1) − (5) h h(4 + h) = h

= l´ım

l´ım−

x < 2, x = 2, x > 2.

h→0−

l´ım (2 + h) = 2

h→0+

f (2 + h) − f (2) 6= h

l´ım

h→0+

f (2 + h) − f (2) h

la derivada no existe en x0 = 2


2. Determine la derivada de las siguientes funciones: 2.1) f (x) =

ln(sen(x + 1)) , √ x−1

Soluci´ on ′

f (x) =

=

ln(sen(x + 1)) √ x−1

√ √ (ln(sen(x + 1)))′ · ( x − 1) − (ln(sen(x + 1))) · ( x − 1)′ √ ( x − 1)2

=

1 sen(x+1)

=

1 sen(x+1)

· (sen(x + 1))

√ · ( x − 1) − (ln(sen(x + 1))) · √ ( x − 1)2

√ · (cos(x + 1)) · ( x − 1) − (ln(sen(x + 1))) · √ ( x − 1)2

1 √ 2 x

2.2) f (x) = (x2 − 1)(x + 5)3 . Soluci´ on

f ′ (x) = (x2 − 1)′ · (x + 5)3 + (x2 − 1) · ((x + 5)3 )′ =

(2x) · (x + 5)3 + (x2 − 1) · (3(x + 5)2 )(x + 5)′

= 2x(x + 5)3 + 3(x2 − 1)(x + 5)2 =

2x(x + 5) + 3(x2 − 1) (x + 5)2

=

5x2 + 10x − 3 (x + 5)2

1 √ 2 x


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