EJERCICIOS RESUELTOS APLICANDO PROPIEDADES DE LA DERIVADA Y OTROS METODOS.
EJERCICIOS N°1:
Comprobar la respuesta de a derivada de las siguientes funciones:
1)
𝑒𝑥 − 1
𝑦 = 𝑒 𝑥 + ln [√ 𝑥
]
𝑒 +1
Resolución:
𝑑
𝑑 𝑥
𝑑
𝑒𝑥 − 1
[𝑦] =
[𝑒 ] +
[ln [√ 𝑥
]]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑒 +1
𝑑
[𝑦] = 𝑒𝑥 +
𝑑𝑥
𝑑
[𝑦] = 𝑒𝑥 +
𝑑𝑥
𝑑
[𝑦] = 𝑒𝑥 +
𝑑𝑥
1
[ (
𝑥
2 𝑒𝑥 + 1
√𝑒𝑥 − 1
[ 𝑒 + 1]
𝑥
𝑥
√𝑒𝑥 − 1 2√𝑒𝑥 − 1
[ 𝑒 + 1 ] [ 𝑒 + 1]
𝑑
[𝑦] = 𝑒𝑥 +
𝑑𝑥
1 𝑒𝑥 − 1 2−1 𝑑 𝑒𝑥 − 1
1
1
1
𝑑
𝑒𝑥 − 1
[√ 𝑥
]
𝑒𝑥 − 1 𝑑𝑥 𝑒 + 1
√
[ 𝑒 𝑥 + 1]
1
[
( 𝑒 𝑥 + 1)
)
]
[
]
𝑑𝑥 𝑒𝑥 + 1
𝑑 𝑥
𝑑
[𝑒 − 1] − (𝑒𝑥 − 1) [𝑒𝑥 + 1]
𝑑𝑥
𝑑𝑥
]
( 𝑒 𝑥 + 1 )2
(𝑒𝑥 + 1)[𝑒𝑥 ] − (𝑒𝑥 − 1)[𝑒𝑥 ]
[
]
( 𝑒 𝑥 + 1 )2
𝑒𝑥 − 1
𝑒𝑥 − 1
√
2√
[ 𝑒𝑥 + 1 ] [ 𝑒𝑥 + 1 ]
1
1
𝑑
1
2𝑒𝑥
[𝑦] = 𝑒𝑥 + [
]
[
]
𝑥
2
𝑒𝑥 − 1
𝑑𝑥
2(
) ( 𝑒 + 1)
𝑒𝑥 + 1
𝑑
𝑒𝑥 + 1
𝑒𝑥
[𝑦] = 𝑒𝑥 + [ 𝑥
][ 𝑥
]
𝑑𝑥
𝑒 − 1 (𝑒 + 1) 2
𝑑
𝑒𝑥 (𝑒𝑥 − 1)(𝑒𝑥 + 1)(𝑒𝑥 + 1) + 𝑒2𝑥 + 𝑒𝑥
[𝑦] =
(𝑒𝑥 − 1)(𝑒𝑥 + 1)(𝑒𝑥 + 1)
𝑑𝑥
𝑑
𝑒𝑥 (𝑒2𝑥 − 1)(𝑒𝑥 + 1) + 𝑒2𝑥 + 𝑒𝑥
[𝑦] =
(𝑒2𝑥 − 1)(𝑒𝑥 + 1)
𝑑𝑥
𝑑
𝑒𝑥 (𝑒3𝑥 + 𝑒2𝑥 − 𝑒𝑥 − 1) + 𝑒2𝑥 + 𝑒𝑥 𝑒4𝑥 + 𝑒3𝑥 − 𝑒2𝑥 − 𝑒𝑥 + 𝑒2𝑥 + 𝑒𝑥
[𝑦] =
=
=
(𝑒3𝑥 + 𝑒2𝑥 − 𝑒𝑥 − 1)
𝑑𝑥
𝑒2𝑥 (𝑒𝑥 + 1) − (𝑒𝑥 + 1)
=
𝑒 4𝑥 + 𝑒 3𝑥
𝑒 3𝑥 (𝑒 𝑥 + 1)
𝑒 3𝑥
=
=
(𝑒 𝑥 + 1)(𝑒 2𝑥 − 1) (𝑒 𝑥 + 1)(𝑒 2𝑥 − 1) 𝑒 2𝑥 − 1
2)
𝑦=
𝑥2
− ln [√1 + 𝑥 2 ]
2
Resolución:
𝑑
1 𝑑 2
𝑑
[𝑦] =
[𝑥 ] −
[ln [√1 + 𝑥2 ]]
𝑑𝑥
2 𝑑𝑥
𝑑𝑥
𝑑
1
1
𝑑
[𝑦] = [2𝑥] − [
] [ √1 + 𝑥 2 ]
𝑑𝑥
2
√1 + 𝑥2 𝑑𝑥
1
𝑑
1
1
1
𝑑
[𝑦] = [2𝑥] − [
] [ (1 + 𝑥2 )2−1 ] [1 + 𝑥2 ]
𝑑𝑥
2
𝑑𝑥
√1 + 𝑥 2 2
𝑑
1
1
[𝑦] = 𝑥 − [
][
] [2𝑥]
𝑑𝑥
√ 1 + 𝑥 2 2√ 1 + 𝑥 2
𝑑
𝑥
[𝑦] = 𝑥 − [
2]
𝑑𝑥
2
(√1 + 𝑥 )
𝑑
𝑥 (1 + 𝑥 2 ) − 𝑥 𝑥 + 𝑥 3 − 𝑥
𝑥3
[𝑦] =
=
=
(1 + 𝑥 2 )
(1 + 𝑥 2 )
𝑑𝑥
1 + 𝑥2
3)
𝑥2 1
𝑦=
− tan−1 (𝑥 2 )
2 2
Resolución:
𝑑
1 𝑑 2
1 𝑑
[𝑦] =
[𝑥 ] −
[tan−1 (𝑥2 )]
𝑑𝑥
2 𝑑𝑥
2 𝑑𝑥
𝑑
1
1
1
𝑑
[𝑦] = [2𝑥] − [
] [𝑥 2 ]
2
2
𝑑𝑥
2
2 1 + (𝑥 ) 𝑑𝑥
𝑑
1
1
[𝑦] = 𝑥 − [
] [2𝑥]
𝑑𝑥
2 1 + 𝑥4
𝑑
𝑥(1 + 𝑥 4 ) − 𝑥 𝑥 + 𝑥 5 − 𝑥
𝑥5
[𝑦] =
=
=
𝑑𝑥
1 + 𝑥4
1 + 𝑥4
1 + 𝑥4
4)
2𝑥𝑦 − 𝑥𝑦 2 = 𝑥 2 𝑦 − 4
Resolución:
2 [[1]𝑦 + [1]𝑥𝑦′ ] − [[1]𝑦 2 + 2𝑦𝑥 𝑦′ ] − [[2𝑥]𝑦 + [1]𝑥 2 𝑦′ ] = 0
2𝑦 + 2𝑥𝑦′ − 𝑦 2 + 2𝑦𝑥𝑦′ − 2𝑥𝑦 + 𝑥 2 𝑦′ = 0
𝑦 ′ [2𝑥 + 2𝑥𝑦 + 𝑥2 ] + 2𝑦 − 𝑦2 − 2𝑥𝑦 = 0
𝑦 ′ [2𝑥 + 2𝑥𝑦 + 𝑥2 ] = −2𝑦 + 𝑦2 + 2𝑥𝑦
𝑦′ =
−2𝑦 + 𝑦2 + 2𝑥𝑦
2𝑥 + 2𝑥𝑦 + 𝑥2
5)
𝑦=
1
3
3
sin(𝑥) cos3 (𝑥) + sin(𝑥) 𝑐𝑜𝑠(𝑥) + 𝑥
4
8
8
Resolución:
𝑑
1 𝑑
3 𝑑
3 𝑑
[𝑦] =
[sin(𝑥) cos3 (𝑥)] +
[sin(𝑥) 𝑐𝑜𝑠(𝑥)] +
[𝑥]
𝑑𝑥
4 𝑑𝑥
8 𝑑𝑥
8 𝑑𝑥
𝑑
1
𝑑
𝑑
3
𝑑
𝑑
3
[𝑦] = [cos3 (𝑥) [sin(𝑥)] + sin(𝑥) [cos3 (𝑥)]] + [cos(𝑥) [sin(𝑥)] + 𝑠𝑖𝑛(𝑥) [𝑐𝑜𝑠(𝑥)]] + [1]
𝑑𝑥
4
𝑑𝑥
𝑑𝑥
8
𝑑𝑥
𝑑𝑥
8
𝑑
1
𝑑
3
3
[𝑦] = [cos3 (𝑥)[cos(𝑥)] + sin(𝑥) [3 cos2 (𝑥)] [cos(𝑥)]] + [cos(𝑥)[cos(𝑥)] + 𝑠𝑖𝑛(𝑥) [− 𝑠𝑖𝑛(𝑥)]] +
𝑑𝑥
4
𝑑𝑥
8
8
𝑑
1
3
3
[𝑦] = [cos3 (𝑥)[cos(𝑥)] + sin(𝑥) [3 cos2 (𝑥)][−sin(𝑥)]] + [cos(𝑥)[cos(𝑥)] + 𝑠𝑖𝑛(𝑥) [− 𝑠𝑖𝑛(𝑥)]] +
𝑑𝑥
4
8
8
𝑑
1
3
3
[𝑦] = [cos4 (𝑥) −3 sin2 (𝑥) [cos2 (𝑥)]] + [cos2(𝑥) − sin2 (𝑥)] +
𝑑𝑥
4
8
8
𝑑
1
3
3
[𝑦] = [cos4 (𝑥) −3(1 − cos2 (𝑥))[cos2(𝑥)]] + [cos2 (𝑥) −(1 − cos2 (𝑥))] +
𝑑𝑥
4
8
8
𝑑
1
1
1
3
3 3
3
[𝑦] = cos4 (𝑥) −3 cos2 (𝑥) + 3 cos4 (𝑥) + cos2 (𝑥) − + cos2 (𝑥) +
𝑑𝑥
4
4
4
8
8 8
8
𝑑
[𝑦] = cos4 (𝑥)
𝑑𝑥
6)
𝑦 = 𝑥 √𝑥[3 ln(𝑥) − 2]
Resolución:
𝑦 = 3𝑥 √𝑥 ln(𝑥) − 𝑥 √𝑥2
𝑑
𝑑
𝑑
[𝑦] = 3. [𝑥√𝑥 ln(𝑥)] − 2. [𝑥√𝑥]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑
𝑑
𝑑
𝑑
𝑑
𝑑
[𝑦] = 3 [√𝑥 ln(𝑥) [𝑥] + 𝑥 ln(𝑥) [√𝑥] + 𝑥√𝑥 [ln(𝑥)]] − 2 [√𝑥 [𝑥] + 𝑥 [√𝑥]]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑
1
1
1
[𝑦] = 3 [√𝑥 ln(𝑥) [1] + 𝑥 ln(𝑥) [
] + 𝑥√𝑥 [ ]] − 2 [√𝑥[1] + 𝑥 [
]]
𝑑𝑥
𝑥
2√ 𝑥
2√𝑥
𝑑
1
1
[𝑦] = 3 [√𝑥 ln(𝑥) + 𝑥 ln(𝑥) [
] + √𝑥 ] − 2 [ √𝑥 + 𝑥 [
]]
𝑑𝑥
2√ 𝑥
2√𝑥
𝑑
1
𝑥
[𝑦] = 3√𝑥 ln(𝑥) + 3𝑥 ln(𝑥) [
] + 3√𝑥 − 2√𝑥 − [ ]
𝑑𝑥
2√ 𝑥
√𝑥
𝑑
1
𝑥
√𝑥
√𝑥
[𝑦] = 3 ( ) √𝑥 ln(𝑥) + 3𝑥 ln(𝑥) [
] + ( ) √𝑥 − [ ]
𝑑𝑥
2√ 𝑥
√𝑥
√𝑥
√𝑥
𝑑
2 𝑥
1
𝑥
𝑥
[𝑦] = 3 ( ) ( ) ln(𝑥) + 3𝑥 ln(𝑥) [
]+( )−[ ]
𝑑𝑥
2 √𝑥
2√ 𝑥
√𝑥
√𝑥
𝑑
6𝑥
3𝑥
[𝑦] = ln(𝑥) (
+
)
𝑑𝑥
2 √ 𝑥 2√ 𝑥
𝑑
9𝑥
[𝑦] = ln(𝑥) (
)
𝑑𝑥
2√ 𝑥
𝑑
9
𝑥
√𝑥
[𝑦] = ln(𝑥) ( ) ( )
𝑑𝑥
2
√𝑥 √𝑥
𝑑
9
[𝑦] = √𝑥 ln(𝑥)
𝑑𝑥
2
7)
𝑥
𝑥
𝑦 = − cot 2 ( ) − 2 ln [sin ( )]
2
2
Resolución:
𝑑
𝑑
𝑥
𝑑
𝑥
[𝑦] = − [cot2 ( )] − 2 [ln [sin ( )]]
𝑑𝑥
𝑑𝑥
2
𝑑𝑥
2
𝑑
𝑥 𝑑
𝑥
1
𝑑
𝑥
[𝑦] = − [2 𝑐𝑜𝑡 ( )] [𝑐𝑜𝑡 ( )] − 2 [
] [sin ( )]
𝑥
𝑑𝑥
2 𝑑𝑥
2
2
sin ( ) 𝑑𝑥
2
𝑑
𝑥
𝑥 𝑑 𝑥
1
𝑥 𝑑 𝑥
[𝑦] = − [2 𝑐𝑜𝑡 ( )] [−𝑐𝑠𝑐2 ( )] [ ] − 2 [
] [cos ( )] [ ]
𝑥
𝑑𝑥
2
2 𝑑𝑥 2
2 𝑑𝑥 2
sin ( )
2
𝑑
𝑥
𝑥
1
1
𝑥
1
[𝑦] = [2 𝑐𝑜𝑡 ( )] [𝑐𝑠𝑐2 ( )] [ ] − 2 [
]
cos
(
)]
[
[
]
𝑥
𝑑𝑥
2
2
2
2
2
sin ( )
2
𝑑
𝑥
𝑥
𝑥
[𝑦] = [𝑐𝑜𝑡 ( )] [𝑐𝑠𝑐2 ( )] − [cot ( )]
𝑑𝑥
2
2
2
𝑑
𝑥
𝑥
[𝑦] = [𝑐𝑜𝑡 ( )] (𝑐𝑠𝑐2 ( ) − 1)
𝑑𝑥
2
2
𝑑
𝑥
𝑥
𝑥
[𝑦] = [𝑐𝑜𝑡 ( )] (𝑐𝑜𝑡2 ( )) = 𝑐𝑜𝑡3 ( )
𝑑𝑥
2
2
2
𝑑
1 𝑒3𝑥 + 9
1 𝑒3𝑥
[𝑦] = ( 3𝑥
) − [ 3𝑥
]
𝑑𝑥
9 𝑒 +9
9 𝑒 +9
𝑑
𝑒3𝑥 + 9 − 𝑒3𝑥
[𝑦] = [
]
𝑑𝑥
9(𝑒3𝑥 + 9)
𝑑
9
1
[𝑦] =
= 3𝑥
3𝑥
𝑑𝑥
9( 𝑒 + 9) 𝑒 + 9
8)
𝑦 = [𝑥]sin(𝑥)
Resolución:
ln(𝑦) = ln([𝑥]sin(𝑥) )
ln(𝑦) = sin(𝑥) ln(𝑥)
ln(𝑦) − sin(𝑥) ln(𝑥) = 0
𝐸𝑥 = ln(𝑦) − sin(𝑥) ln(𝑥)
Derivada parcial con respecto a x:
𝑑
𝑑
𝑑
[𝐸𝑥 ] = 𝐸𝑥′ =
[ln(𝑦)] −
[sin(𝑥) ln(𝑥)]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑
𝑑
𝑑
[𝐸𝑥 ] = 𝐸𝑥′ = − [ln(𝑥) [sin(𝑥)] + sin(𝑥) [ln(𝑥)]]
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑
1
[𝐸𝑥 ] = 𝐸𝑥′ = − [ln(𝑥)[cos(𝑥)] + sin(𝑥) [ ]]
𝑑𝑥
𝑥
𝐸𝑦 = ln(𝑦) − sin(𝑥) ln(𝑥)
Derivada parcial con respecto a y:
𝑑
𝑑
𝑑
[ln(𝑦)] −
[sin(𝑥) ln(𝑥)]
[𝐸𝑦 ] = 𝐸𝑦′ =
𝑑𝑦
𝑑𝑦
𝑑𝑦
𝑑
1
[𝐸𝑦 ] = 𝐸𝑦′ = [ ]
𝑑𝑦
𝑦
1
− [ln(𝑥)[cos(𝑥)] + sin(𝑥) [𝑥]]
𝐸𝑥′
𝑦 =− ′ =−
=
1
𝐸𝑦
′
𝑦
[
=
=
x ln(𝑥)[cos(𝑥)] + sin(𝑥)
]
𝑥
=
1
𝑦
𝑦[x ln(𝑥)[cos(𝑥)] + sin(𝑥)]
𝑥
= [𝑥]sin(𝑥) [ ln(𝑥)[cos(𝑥)] +
sin(𝑥)
𝑥
9)
𝑥 cos(𝑦) + 𝑦 sin(𝑥) = 𝑥 2 𝑦 − 4
Resolución:
=
]
[[1] cos(𝑦) + 𝑥[− sin(𝑦) 𝑦′ ]] + [sin(𝑥) [1]𝑦′ + 𝑦 cos(𝑥)] = [2𝑥𝑦 + [1]𝑦′ ]
cos(𝑦) − 𝑥[sin(𝑦) 𝑦′ ] + sin(𝑥) 𝑦′ + 𝑦 cos(𝑥) − 2𝑥𝑦 − 𝑦′ = 0
𝑦 ′ [−𝑥[sin(𝑦)] + sin(𝑥) − 1] + cos(𝑦) + 𝑦 cos(𝑥) − 2𝑥𝑦 =
𝑦 ′ [−𝑥[sin(𝑦)] + sin(𝑥) − 1] = − cos(𝑦) − 𝑦 cos(𝑥) + 2𝑥𝑦
𝑦 ′ [(−1)(𝑥[sin(𝑦)] − sin(𝑥) + 1)] = −[cos(𝑦) + 𝑦 cos(𝑥) − 2𝑥𝑦]
𝑦′ =
cos(𝑦) + 𝑦 cos(𝑥) − 2𝑥𝑦
𝑥[sin(𝑦)] − sin(𝑥) + 1
