Regla de la cadena

5. Regla de la cadena

Ejercicio 6.1

Calcule la derivada de las siguientes funciones

1: $f (y) = {[{(y^{2} + 3)}^{4} - 1]}^{3}$
$f^{'} (y) = 24 {[{(y^{2} + 3)}^{4} - 1]}^{2} \cdot {(y^{2} + 3)}^{3} \cdot y$ .
2: $f (𝜃) = sen (5 𝜃) {sec}^{2} (5 𝜃)$
$f^{'} (𝜃) = 5 sec (5 𝜃) + 10 {sen}^{2} (5 𝜃) {sec}^{3} (5 𝜃)$ .
3: $f (x) = {(\frac{x - 7}{x + 2})}^{3}$
$f^{'} (x) = 27 {(x - 7)}^{2} ∕ {(x + 2)}^{4}$ .
4: $f (t) = \frac{3 t^{4}}{\sqrt{t^{2} - 5 t}}$
$f^{'} (t) = \frac{18 t^{5} - 105 t^{4}}{2 {(t^{2} - 5 t)}^{3 ∕ 2}}$ .
5: $f (x) = \frac{{tan}^{3} (x) - 1}{sen (x) + 2}$
$f^{'} (x) = \frac{3 {tan}^{2} (x) {sec}^{2} (x) (sin (x) + 2) - cos (x) ({tan}^{3} (x) - 1)}{{(sin (x) + 2)}^{2}}$
6: $f (x) = e^{2 x^{2} - 5 x + 3}$
$f^{'} (x) = (4 x - 5) e^{2 x^{2} - 5 x + 3}$ .
7: $f (q) = {(q - 3 e^{q ∕ 3})}^{5}$
$f^{'} (q) = 5 {(q - 3 e^{\frac{q}{3}})}^{4} (1 - e^{\frac{q}{3}})$ .
8: $f (x) = \frac{e^{x} - 1}{e^{x} + 1}$
$f^{'} (x) = \frac{2 e^{x}}{{(e^{x} + 1)}^{2}}$ .
9: $f (t) = ln (t \sqrt{t^{2} + 1})$
$f^{'} (t) = t^{- 1} + t {(t^{2} + 1)}^{- 1} .$
10: $f (p) = ln [\frac{(p^{3} - 1) e^{- p^{2}}}{\sqrt{1 - 5 p}}]$
$f^{'} (p) = 3 p^{2} {(p^{3} - 1)}^{- 1} - 2 p + 5 {(2 - 10 p)}^{- 1}$ .
11: $f (z) = \sqrt{1 + ln z} + ln (1 + \sqrt{z})$
$f^{'} (z) = \frac{1}{2} z^{- 1} {(1 + ln z)}^{\frac{- 1}{2}} + \frac{1}{2} z^{\frac{- 1}{2}} {(1 + \sqrt{z})}^{- 1}$ .
12: $f (x) = e^{3 x} g ({ln}^{2} x)$ , donde $g$ es derivable.
$f^{'} (x) = 3 e^{3 x} g ({ln}^{2} x) + 2 e^{3 x} g^{'} ({ln}^{2} x) \frac{ln x}{x} .$
13: $f (y) = {ln}^{2} (2 y + 6)$
$f^{'} (y) = \frac{2 ln (2 y + 6)}{(y + 3)} .$
14: $f (z) = {ln}^{2} [ln (2 z^{3} - 8 z)]$
$f^{'} (z) = \frac{2 ln (ln (2 z^{3} - 8 z)) (6 z^{2} - 8)}{[(2 z^{3} - 8 z) ln (2 z^{3} - 8 z)]} .$
15: $f (x) = \frac{e^{x} + tan (x + 1)}{sen (2 x)}$
$f^{'} (x) = \frac{sen (2 x) (e^{x} + {sec}^{2} (x + 1)) - 2 cos (2 x) (e^{x} + tan (x + 1))}{{sen}^{2} (2 x)}$
16: $f (w) = ln (\frac{\sqrt{w - 1}}{w^{3} cos (w^{2})})$
$f^{'} (w) = \frac{1}{2 (w - 1)} - \frac{3}{w} + \frac{sen (w^{2}) \cdot 2 w}{cos (w^{2})}$ .
17: $h (z) = {arccos}^{2} (\frac{e^{- z}}{z})$
$h^{'} (z) = 2 arccos (\frac{e^{- z}}{z}) \cdot \frac{- 1}{\sqrt{1 - {(\frac{e^{- z}}{z})}^{2}}} \cdot \frac{- z e^{- z} - e^{- z}}{z^{2}}$
18: $f (x) = {arctan}^{3} (ln (x^{2} + e^{x}))$
$f^{'} (x) = \frac{3 {arctan}^{2} (ln (x^{2} + e^{x})) \cdot (2 x + e^{x})}{(x^{2} + e^{x}) \cdot (1 + {ln}^{2} (x^{2} + e^{x}))} .$
19: $h (x) = ln (3^{g (x^{3} + 4)} + 1)$
$h^{'} (x) = \frac{3 ln 3 \cdot 3^{g (x^{3} + 4)} \cdot g^{'} (x^{3} + 4) \cdot x^{2}}{3^{g (x^{3} + 4)} + 1} .$
20: $f (x) = \frac{\sqrt{1 - 2 x^{2}} (x + 3)}{{(4 x - 1)}^{2}}$
$f^{'} (x) = \frac{\sqrt{1 - 2 x^{2}} (x + 3)}{{(4 x - 1)}^{2}} (\frac{- 2 x}{1 - 2 x^{2}} + \frac{1}{x + 3} - \frac{8}{4 x - 1})$ .
21: $f (v) = \sqrt{\frac{{(v - 1)}^{3}}{(2 v + 7) {(5 - v)}^{2}}}$
$f^{'} (v) = \sqrt{\frac{{(v - 1)}^{3}}{(2 v + 7) {(5 - v)}^{2}}} (\frac{3}{2 v - 2} - \frac{1}{2 v + 7} + \frac{1}{5 - v})$
22: $f (t) = t^{ln t}$
$f^{'} (t) = 2 t^{ln t - 1} ln t .$
23: $f (w) = e^{w} w^{1 - w}$
$f^{'} (w) = e^{w} w^{1 - w} (w^{- 1} - ln w)$ .
24: $g (z) = sec (e^{1 - 2 z})$
$g^{'} (z) = - 2 e^{1 - 2 z} sec (e^{1 - 2 z}) tan (e^{1 - 2 z})$ .
25: $h (u) = e^{u sen u} + {ln}^{3} (3 - 2 u^{2})$
$h^{'} (u) = e^{u sen u} (sen u + u cos u) - \frac{12 u {ln}^{2} (3 - 2 u^{2})}{3 - 2 u^{2}}$ .
26: $h (u) = {cos}^{4} (sen (k u^{3}))$
$h^{'} (u) = - 12 k u^{2} {cos}^{3} (sen (k u^{3})) sen (sen (k u^{3})) cos (k u^{3})$ .
27: $g (u) = {ln}^{2} (sen u) + ln (1 - e^{2 u})$
$g^{'} (u) = 2 cot u ln (sen u) - \frac{2 e^{2 u}}{1 - e^{2 u}}$ .
28: $g (u) = ln (sec (u^{3}) + tan (u^{3}))$
$g^{'} (u) = 3 u^{2} sec (u^{3})$ .

Ejercicio 6.2

f

es una función derivable, obtenga la derivada de las siguientes funciones

1: $y = \frac{f (ln z)}{z e^{z}}$
$y^{'} = \frac{f^{'} (ln z) - (1 + z) f (ln z)}{z^{2} e^{z}}$ .
2: $y = f (e^{- u}) e^{f (u)}$
$y^{'} = e^{f (u)} [f (e^{- u}) f^{'} (u) - e^{- u} f^{'} (e^{- u})]$ .
3: $y = f (w^{2 n}) - {[f (w)]}^{n}$
$y^{'} = 2 n w^{2 n - 1} f^{'} (w^{2 n}) - n {[f (w)]}^{n - 1} f^{'} (w) .$
4: $y = e^{4 w} f ({ln}^{3} w)$
$y^{'} = e^{4 w} [4 f ({ln}^{3} w) + \frac{3 {ln}^{2} w}{w} f^{'} ({ln}^{3} w)]$ .

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