This document provides solutions to calculating the derivative functions of various given functions. It includes: 1) Finding the derivative functions of polynomials, rational functions, exponential functions, logarithmic functions, trigonometric functions, and composite functions. 2) The solutions provide the step-by-step work and final derivative function for each problem. 3) There are over 25 problems covered across multiple pages with the aim of teaching calculation of derivative functions.

1. CÁLCULO DE DERIVADAS
1
Halla la función derivada de las siguientes funciones:
Sol. y′ = 6x − 6
1
1
Sol. y′ =
+
3 2
2 x 3 x
−3
Sol. y′ = 2
2x x
Sol. y′ = 6x 2 + 6x
1
Sol. y′ =
3
3x 2
1
Sol. y′ =
+ 3x −
2
2
a) y = 3x 2 − 6 x + 5
b) y = x + 3 x
1
c) y =
x x
d) y = 2x 3 + 3x 2 − 6
x
e) y = + 2
3
x3 3 2 x
+ x −
f) y =
2 2
2
2
Halla la función derivada de las siguientes funciones:
a) y = sen x cos x
Sol. y′ = cos2 x − sen2 x = cos 2x
b) y = x e x
Sol. y′ = e x (1 + x )
c) y = x 2x
Sol. y′ = 2x (1 + x ln 2 )
(
)
Sol. y′ = 2x ln x +
d) y = x 2 + 1 ln x
(
)
(
)
(
Sol. y′ = 15x 4 − 1 log2 x + 3x 5 − x
e) y = 3x 5 − x log2 x
x2 + 1
x2 − 1
Sol. y′ =
g) y =
x 3 + 3x 2 − 5x + 3
x
h) y =
log x
x
1
) x ln 2
Sol. y′ =
f) y =
3
x2 + 1
x
−4x
(x
2
)
−1
2
2x 3 + 3x 2 − 3
x2
1
− log x
loge − log x 1
e
ln10
Sol. y′ =
=
= 2 log
2
2
x
x
x
x
Halla la función derivada de las siguientes funciones:
1
5
a) y = 2x + 3 5x
Sol. y′ =
+
3
2x 3 25x 2
b) y = sen x 2 − 5x + 7
Sol. y′ = ( 2x − 5 ) cos x 2 − 5x + 7
(
c) y = 3 ( 5 x + 3 )
)
2
(
Sol. y′ =
)
10
d) y = sen ( 3x + 1) cos ( 3x + 1)
3 5x + 3
Sol. y′ = 3 cos ( 6x + 2 )
log x 2
e) y =
x
f) y = cos ( 3x − π )
2
2
− log x 2
− 2log x
2
e
ln10
′ = ln10 2
Sol. y
=
= 2 log
2
x
x
x
x
Sol. y′ = −3sen ( 3x − π )
g) y = 1 + 2x
Sol. y′ =
h) y = x e
2x +1
3
1
1 + 2x
′ = e2 x +1 (1 + 2x )
Sol. y
1/7

2. i) y =
(
)
sen x 2 + 1
Sol. y′ =
1 − x2
j) y = x sen (π − x )
k) y =
Sol. y′ =
⎛ 3x
⎞
− 5⎟
l) y = ⎜
⎝ 2
⎠
3x − 7
m) y =
x+2
)
(
)
(
)
(
1 − x2 1 − x2
)
Sol. y′ = sen (π − x ) − x cos (π − x )
1
x2
3
(
2x 1 − x 2 cos x 2 + 1 + xsen x 2 + 1
4
−2
3x 3 x 2
⎛ 3x
⎞
Sol. y′ = 6 ⎜
− 5⎟
⎝ 2
⎠
13
Sol. y′ =
2
( x + 2)
3
n) y = e − x ln ( cos x )
Sol. y′ = −e − x ⎡ln ( cos x ) + tg x ⎤
⎣
⎦
ñ) y = tg x
Sol. y′ =
⎛ x + 1⎞
o) y = ln ⎜
⎟
⎝ x ⎠
Sol. y′ =
⎛ x ⎞
p) y = ⎜
⎟
⎝ x + 1⎠
1
2
2 cos x tg x
−1
x +x
2
5
Sol. y′ =
5x 4
( x + 1)
6
2
⎛ 1 − x ⎞3
q) y = ⎜
⎟
⎝ 1+ x ⎠
Sol. y′ =
x2 − 4
x
r) y =
Sol. y′ =
2
4
x
x +1
(
3 (1 + x ) 3 (1 − x )(1 + x )
Sol. y′ =
x −1
s) y =
−4
2
x2 − 4
1
( x + 1)
7
( 2 x − 3)
x
2
x
t) y = 2 x − 3
)
7
Sol. y′ =
6
4
(Ejercicio resuelto) Halla la función derivada de las siguientes funciones:
1
cos x
⋅ cos x; y′ =
a) y = sen x → y′ =
2 sen x
2 sen x
y′ = cos x 2 + 5x − 1 ⋅
b) y = sen x 2 + 5x − 1 →
y′ =
( 2x + 5 ) cos
1
2 x 2 + 5x − 1
⋅ ( 2x + 5 )
x 2 + 5x − 1
2 x 2 + 5x − 1
⋅ ( 2x + 3 ) ;
+3 x
→
y′ = 5 e x
d) y = ln ( ln x )
→
y′ =
1 1
⋅ ;
ln x x
)
→
c) y = 5 e x
(
2
e) y = x 2 − x + sen x
4
2
+3 x
y′ =
y′ = 5 ( 2x + 3 ) e x
2
+3 x
1
x ln x
(
)
3 ⎛
1
⎞
y′ = 4 x 2 − x + sen x ⎜ 2x −
+ cos x ⎟
2 x
⎝
⎠
2/7

3. f)
y = x x +1
⎛ 2 ( x + 1) + x ⎞
⋅⎜
⎜ 2 x + 1 ⎟;
⎟
3
2
⎠
x +x ⎝
2
g) y =
4 4 x 2 ( x + 1)
1− x
1+ x
⋅
3x + 2
2 4 x3 + x 2 2 x + 1
=
1
⋅
3x + 2
2 4 x 2 ( x + 1) 2 4 ( x + 1)2
1− x
1+ x
3
y′ =
→
−1 ⋅ (1 + x ) − (1 − x ) ⋅ 1
(1 + x )
y′ =
→
−1
y′ ==
(1 + x )
i)
1
y′ =
3x + 2
h) y =
y′ =
1 ⎞
⎛
⋅ ⎜ 1⋅ x + 1 + x ⋅
⎟
2 x +1⎠
2 x x +1 ⎝
1
1
y′ =
y′ =
y′ =
→
2
1− x
2
1+ x
−1
=
1− x
1+ x
⋅
1
(1 + x ) (1 − x )
y′ =
→
−1 − x − 1 + x
2
;
y′ =
1
1− x
2
1+ x
−2
(1 + x )
⋅
2
−2
(1 + x )
2
−1
(1 + x ) (1 − x )
3
1+ x
(1 + x ) (1 + x )(1 − x )
1− x
1+ x
=
4
−1
y = ln
=
(1 + x )
−1⋅ (1 + x ) − (1 − x ) ⋅ 1
⋅
=
2
(1 + x )
2
;
y′ =
−1
(1 + x )
1 − x2
−2
−2
1
−2
;
⋅
=
=
1 − x (1 + x )2 (1 − x )(1 + x ) 2 (1 − x )(1 + x )
1+ x
1+ x
y′ =
−2
1 − x2
De otra forma: Desarrollando el logaritmo previamente:
1
1
−1
1
y = ln (1 − x ) − ln (1 + x ) → y′ =
.1 =
⋅ ( −1) −
−
1− x
1+ x
1− x 1+ x
−1 − x − (1 − x ) −1 − x − 1 + x
−2
y′ =
; y′ =
=
1 − x2
(1 − x )(1 + x ) (1 − x )(1 + x )
j)
y=
y′ =
k) y =
1 − tg x
1 + tg x
−1 − tg x − 1 + tg x
−1
1
⋅ (1 + tg x ) − (1 − tg x ) ⋅
2
2
cos x =
cos2 x
y′ = cos x
2
2
(1 + tg x )
(1 + tg x )
→
−2
cos x (1 + tg x )
2
1 − tg x
1 + tg x
→
2
y′ =
⎛ 1 − tg x ⎞
⋅D⎜
⎟
1 − tg x
⎝ 1 + tg x ⎠
2
1 + tg x
1
⎛ 1 − tg x ⎞
−2
En el ejercicio anterior se calculó: D ⎜
⎟=
2
2
⎝ 1 + tg x ⎠ cos x (1 + tg x )
1
y′ =
2
⋅
−2
1 − tg x cos x (1 + tg x )
1 + tg x
2
2
−1
=
(1 + tg x ) (1 − tg x )
4
2
cos x
3/7
1 + tg x

4. −1
y′ ==
cos2 x
y′ =
l)
=
−1
cos x (1 + tg x )
2
(1 + tg x )(1 − tg x )
−1
cos2 x (1 + tg x ) 1 − tg2 x
→
y′ =
+
ln 3 x2 1
⋅3 ;
2
y′ =
y = 3 x +1
y′ =
(1 + tg x ) (1 − tg x )
3
1
2 3 x +1
⋅ 3 x +1 ⋅ ln3 =
+
1 3 x +1
ln 3 3 x +1 ln 3 ( x +1) − x2 1
⋅
⋅ ln 3 =
⋅ x +1 =
⋅3
2 3 x +1
2
2
32
ln 3 x +1
3
2
m) y = log ( sen x ⋅ cos x ) ;
y = 2 ⋅ log ( sen x ⋅ cos x )
2
1
1
⋅
⋅ ( cos x ⋅ cos x + sen x ( - sen x ) )
sen x ⋅ cos x ln10
2 cos2 x − sen2 x
2
y′ =
⋅ cos2 x − sen2 x =
ln10 ⋅ sen x ⋅ cos x
ln10 ⋅ sen x ⋅ cos x
Teniendo en cuenta que: cos 2x = cos 2 x - sen 2 x y que
1
sen 2x = 2 sen x cos x → sen x cos x = sen 2x
2
2 ⋅ cos 2x
2 ⋅ cos 2x
4 ⋅ cos 2x
4 cos 2x
=
=
⋅
y′ =
; y′ =
sen 2x ln10 ⋅ sen 2x ln10 ⋅ sen 2x
ln10 sen 2x
ln10 ⋅
2
2
cos 2x
1
Teniendo en cuenta que:
= cotg 2x =
, se tiene:
sen 2x
tg 2x
4
4
y′ =
⋅ cotg2x; y′ =
ln10
ln10 ⋅ tg2x
→
y′ = 2 ⋅
(
)
(
)
De otra forma:
⎛ sen 2x ⎞
y = 2 ⋅ log ⎜
⎟
⎝ 2 ⎠
1
1
1
4 cos 2x
4
y′ = 2 ⋅
=
⋅ cotg2x
⋅
⋅ ⋅ ( cos 2x ) ⋅ 2 =
sen 2x ln10 2
ln10 sen 2x ln10
2
y = log ( sen x ⋅ cos x ) ;
2
→
y = 2 ⋅ log ( sen x ⋅ cos x ) ;
n) y = sen2 x + cos2 x + x → y′ = 2 sen x cos+ 2cos x ( − sen x ) + 1
y′ = 2 sen x cos x − 2 sen x cos x + 1; y′ = 1
ñ) y = sen x + 1 ⋅ cos x − 1
1
1
→ y′ = cos x + 1 ⋅
⋅ cos x − 1 + sen x + 1 ⋅ − sen x − 1 ⋅
2 x +1
2 x −1
cos x + 1 ⋅ cos x − 1 sen x + 1 ⋅ sen x − 1
y′ =
−
2 x +1
2 x −1
(
(
)
o) y = sen 3x 5 − 2 x + 3 2x ;
(
y = sen 3x 5 − 2 x + 3 2 ⋅ 3 x
)
)
⎛
2
1 ⎞
y′ = cos 3x 5 − 2 x + 3 2x ⋅ ⎜ 15x 4 −
+32⋅
⎟
2 x
3 3 x2 ⎠
⎝
3
⎛
1
2 ⎞
5
3
y′ = ⎜ 15x 4 −
+
⎟ ⋅ cos 3x − 2 x + 2x
⎜
3 2 ⎟
x 3 x ⎠
⎝
→
(
)
(
)
4/7

5. y′ =
p) y = sen x + x 2 + 1 →
2
y′ =
2 sen x + x + 1
2
)
y = cos2 3 x + 9 − 6x + x 2 ;
)
(
3
3
(
2 sen x + x 2 + 1
y = cos2 3 x 2 − 5x + 9
(x
(
)
2
− 5x + 9
)
2
⋅ ( 2x − 5 )
2
− ( 2x − 5 ) ⋅ 2 ⋅ sen 3 x 2 − 5x + 9 ⋅ cos 3 x 2 − 5x + 9
(
y′ =
sen2a = 2 sena cosa →
)
(
− ( 2x − 5 ) ⋅ sen 2 3 x 2 − 5x + 9
(
( 5 − 2x ) ⋅ sen ( 2 3 x 2 − 5x + 9 )
(
3 3 x 2 − 5x + 9
y = arc sen x
→
)
→
1
y′ =
y′ =
( x)
2
−1
y′ =
1−
2
( x)
1
⋅
1
2 x
1
2 x
1
;
y′ =
;
y′ =
y′ =
1
⋅
1
⋅
1
1− x 2 x
−1
;
1− x 2 x
1
1
⋅
;
1+ x 2 x
y = arc tg x
5
y′ =
y′ =
y′ =
b) y =
( x)
5
Sol. y′ =
x2 − 4
x2
Sol. y′ =
d) y = x 2 sen x + 2x cos x − 2 sen x
e) y = tg2 e3x
2
x
2
y′ = 2x arc cos + x 2
x
5x 4
( x + 1)
6
−x2 + 8
Sol. y′ = x 2 cos x
Sol. y′ = 6 e3 x tg e3 x sec 2 e3 x
y = x 2 arc cos
−1
−2
;
2
2
⎛2⎞ x
1− ⎜ ⎟
⎝x⎠
5/7
2 x − x2
−1
2 x − x2
2 x (1 + x )
x3 x2 − 4
Sol. y′ = sec x
c) y = ln ( sec x + tg x )
f)
2 x
;
1
1
Halla la función derivada de las siguientes funciones:
1+
2
⋅
2
⋅
t)
⎛ x ⎞
a) y = ⎜
⎟
⎝ x + 1⎠
→
)
)
2
1−
s) y = arc cos x
)
2
3 3 x 2 − 5x + 9
r)
cos x + 2x
−2 ( 2x − 5 ) ⋅ cos 3 x 2 − 5x + 9 ⋅ sen 3 x 2 − 5x + 9
3 3 x 2 − 5x + 9
y′ =
y′ =
1
y′ = 2 cos 3 x 2 − 5x + 9 ⋅ − sen 3 x 2 − 5x + 9 ⋅
3 3 x 2 − 5x + 9
y′ =
⋅ ( cos x + 2x ) ;
(
q) y = cos2 3 x + ( 3 − x ) ;
→
1

6. y′ = 2x arc cos
2
+
x
2
4
1− 2
x
⎛
2
y′ = 2x ⎜ arc cos +
x
⎝
y′ = 2x arc cos
;
2
+
x
2x
x2 − 4
;
⎞
⎟
x −4⎠
1
2
g) y = ( x − 1) 2x − x 2 + arcsen ( x − 1)
2 − 2x
y′ = 1⋅ 2x − x 2 + ( x − 1) ⋅
y′ = 2x − x 2 + ( x − 1)
y′ = 2x − x 2 +
y′ =
y′ =
2 2x − x
2 (1 − x )
2 2x − x
x − x2 − 1 + x
2x − x
2
2x − x 2
(
2 2x − x 2
(
)
2x − x 2
2x − x
2
)
2
=
1 − ( x − 1)
1 − x + 2x − 1
2x − x 2
2 2x − x
2
)
;
;
4x − 2x 2
(
;
2
2x − x 2
=
2
1
+
1
+
2x − x 2 + x − x 2 − 1 + x + 1
2
1
+
2
=
(
2 2x − x 2
2x − x
2x − x
2
);
2
;
2x − x 2
y′ = 2 2x − x 2
(
)
x
− 2x
2
1
+x⋅
⋅ 2x +
4 + x2
h) y = x ln 4 + x 2 + 4 arc tg
(
y′ = ln 4 + x 2
(
)
(
)
y′ = ln 4 + x 2
(
)
y′ = ln ( 4 + x )
y′ = ln 4 + x 2
)
4
1
⋅ − 2;
2
x 2
1+
4
2x 2
16
1
+
+
⋅ − 2;
4 + x2 4 + x2 2
2x 2
8
+
+
− 2;
2
4+x
4 + x2
2x 2 + 8 − 8 − 2x 2
0
;
+
= ln 4 + x 2 +
2
4+x
4 + x2
y′ = 1⋅ ln 4 + x 2
(
)
2
6
Aplica la derivación logarítmica para derivar:
a) y = x 3x
Se toma logaritmo neperiano en cada miembro:
ln y = ln x 3x ; ln y = 3x ln x
Se deriva en cada miembro:
1
1
y′
⋅ y′ = 3 ⋅ ln x + 3 x ⋅ ;
= 3 ⋅ ln x + 3
D ( ln y ) = D ( 3x ln x ) ;
y
y
x
Se despeja la función derivada de y: y′ = ( 3ln x + 3 ) y
Se sustituye la función y por su expresión:
y′ = ( 3 ln x + 3 ) x 3x ;
y′ = 3x 3 x ( ln x + 1)
6/7

7. x + 1 ⎞ x +1
⎛
x
Sol. y′ = ⎜ ln x +
x ⎟
⎝
⎠
b) y = x x +1
1⎞
⎛
c) y = ⎜ 1 + ⎟
x⎠
⎝
d) y = ( ln x )
x
x +1
⎛ sen x ⎞
e) y = ⎜
⎟
⎝ x ⎠
f)
y = x tg x
⎡ ⎛
1⎞
1 ⎤⎛
1⎞
Sol. y′ = ⎢ln ⎜ 1 + ⎟ −
⎥ ⎜1 + x ⎟
x ⎠ 1+ x ⎦ ⎝
⎠
⎣ ⎝
x + 1⎤
x +1
⎡
ln x )
Sol. y′ = ⎢ln ( ln x ) +
x ⎥(
ln x ⎦
⎣
x
x
⎡ ⎛ sen x ⎞
⎤ ⎛ sen x ⎞
Sol. y′ = ⎢ln ⎜
⎟ + x cotg x − 1⎥ ⎜ x ⎟
⎠
⎣ ⎝ x ⎠
⎦⎝
tg x ⎞ tg x
⎛ ln x
Sol. y′ = ⎜
+
x
2
x ⎟
⎝ cos x
⎠
7/7
x