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Derivada de una función - TEC-Digital

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42 Capítulo 2: <strong>Derivada</strong>s<br />

Ejercicios.<br />

Determine f ′ (x) si<br />

a. f(x) = cot(5 x )<br />

b. f(x) = 2 3√ cot x<br />

c. f(x) = cot(5x 2 + 5 ln x)<br />

5. Dx(sec x) = sec x tan x, x = (2n + 1) π<br />

, n ∈ Z<br />

2<br />

Prueba: Ejercicio para el estudiante.<br />

Si u = g(x), aplicando la regla <strong>de</strong> la ca<strong>de</strong>na se obtiene que Dx(sec u) = sec u tan u Dxu.<br />

Ejemplo 5<br />

a. Dx[sec(2x 2 )] = sec(2x 2 ) tan(2x 2 )Dx(2x 2 ) = 4x sec(2x 2 ) tan(2x 2 )<br />

b. Dx(e sec x ) = e sec x sec x tan x<br />

c. Dx sec<br />

Ejercicios.<br />

<br />

2<br />

x<br />

= sec<br />

<br />

2<br />

x<br />

Determine f ′ (x) si<br />

<br />

2x − 4<br />

a. f(x) = sec<br />

x<br />

b. f(x) = sec 3√ x 2 + 1<br />

c. f(x) = 3x<br />

sec 4x<br />

tan<br />

<br />

2<br />

x<br />

Dx<br />

6. Dx[csc x] = − csc x cot x, x = nπ, n ∈ Z.<br />

Prueba: Ejercicio para el estudiante<br />

<br />

2<br />

x<br />

= −2<br />

sec<br />

x2 <br />

2<br />

tan<br />

x<br />

<br />

2<br />

x<br />

x = 0<br />

Si u = g(x), aplicando la regla <strong>de</strong> la ca<strong>de</strong>na se obtiene que Dx(csc u) = − csc u cot u Dxu.<br />

Ejemplo 6<br />

a. Dx[csc(2 + x 2 )] = − csc(2 + x 2 ) cot(2 + x 2 ) Dx(2 + x 2 ) = −2x csc(2 + x 2 ) cot(2 + x 2 )<br />

b. Dx[csc(2 x )] = − csc 2 x cot 2 x Dx2 x = − csc 2 x cot 2 x ln 2 = −2x ln 2 csc 2 x cot 2 x<br />

c. Dx ln (csc x) = 1<br />

· (− csc x cot x) = − cot x<br />

csc x

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