Exercicis de derivades destinat a batxillerat.
Per treballar autònomament es pot fer servir aquestes dues eines:
Per enganxar una fórmula s'ha de activar editar i buscar al fórmula, seleccionar-la i enganxar-la a la web Symbolab, no seleccioneu les dues capsules seguents <math></math> pròpies de "wikipedia", només lo que està dins d'elles. <math>\left(frac{1}{x}+x^3-4x^2+5\right)'</math> També es pot utilitzar l'aplicació Wolfram Alpha que fa el mateix. També es pot utilitzar l'aplicació Photomath que permet amb una càmera resoldre problemes enunciats físicament que es puguin veure. Linealitat de la derivada [ edit ] Multiplicació de funcions [ edit ] A partir d'ara per aplicar els mètodes omplireu el requadre fg que teniu a la dreta.
f
(
x
)
=
{\displaystyle f(x)=}
f
′
(
x
)
=
{\displaystyle f\,'(x)=}
g
(
x
)
=
{\displaystyle g(x)=}
g
′
(
x
)
=
{\displaystyle g\,'(x)=}
(
f
(
x
)
⋅
g
(
x
)
)
′
{\displaystyle \left(f(x)\cdot g(x)\right)'}
=
f
′
(
x
)
⋅
g
(
x
)
+
f
(
x
)
⋅
g
′
(
x
)
{\displaystyle =f\,'(x)\cdot g(x)+f(x)\cdot g\,'(x)}
Exemples
Divisió de funcions [ edit ]
f
(
x
)
=
{\displaystyle f(x)=}
f
′
(
x
)
=
{\displaystyle f\,'(x)=}
g
(
x
)
=
{\displaystyle g(x)=}
g
′
(
x
)
=
{\displaystyle g\,'(x)=}
(
f
(
x
)
g
(
x
)
)
′
{\displaystyle \left({\frac {f(x)}{g(x)}}\right)'}
=
f
′
(
x
)
⋅
g
(
x
)
−
f
(
x
)
⋅
g
′
(
x
)
g
(
x
)
2
{\displaystyle ={\frac {f\,'(x)\cdot g(x)-f(x)\cdot g\,'(x)}{g(x)^{2}}}}
Exemples
2)
h
(
x
)
=
x
4
−
1
x
{\displaystyle h(x)={\frac {x^{4}-1}{x}}}
h
′
(
x
)
=
(
x
4
−
1
x
)
′
=
{\displaystyle h'(x)=\left({\frac {x^{4}-1}{x}}\right)'=}
(
4
x
3
)
⋅
x
−
(
x
4
−
1
)
⋅
1
x
2
=
{\displaystyle {\frac {\left(4x^{3}\right)\cdot x-\left(x^{4}-1\right)\cdot 1}{x^{2}}}=}
4
x
4
−
x
4
+
1
x
2
=
{\displaystyle {\frac {4x^{4}-x^{4}+1}{x^{2}}}=}
3
x
4
+
1
x
2
{\displaystyle {\frac {3x^{4}+1}{x^{2}}}}
3)
h
(
x
)
=
x
+
x
x
{\displaystyle h(x)={\frac {{\sqrt {x}}+x}{x}}}
h
′
(
x
)
=
(
x
+
x
x
)
′
=
{\displaystyle h'(x)=\left({\frac {{\sqrt {x}}+x}{x}}\right)'=}
(
1
2
x
+
1
)
⋅
x
−
(
x
+
x
)
⋅
1
x
2
=
{\displaystyle {\frac {\left({\frac {1}{2{\sqrt {x}}}}+1\right)\cdot x-\left({\sqrt {x}}+x\right)\cdot 1}{x^{2}}}=}
x
2
x
+
x
−
x
−
x
x
2
=
{\displaystyle {\frac {{\frac {x}{2{\sqrt {x}}}}+x-{\sqrt {x}}-x}{x^{2}}}=}
x
⋅
x
2
x
⋅
x
+
x
−
x
−
x
x
2
=
{\displaystyle {\frac {{\frac {x\cdot {\sqrt {x}}}{2{\sqrt {x}}\cdot {\sqrt {x}}}}+x-{\sqrt {x}}-x}{x^{2}}}=}
1
2
x
+
x
−
x
−
x
x
2
=
{\displaystyle {\frac {{\frac {1}{2}}{\sqrt {x}}+x-{\sqrt {x}}-x}{x^{2}}}=}
(
1
2
−
1
)
x
+
x
−
x
x
2
=
{\displaystyle {\frac {\left({\frac {1}{2}}-1\right){\sqrt {x}}+x-x}{x^{2}}}=}
−
1
2
x
x
2
=
{\displaystyle {\frac {-{\frac {1}{2}}{\sqrt {x}}}{x^{2}}}=}
−
1
2
x
x
2
{\displaystyle -{\frac {1}{2}}{\frac {\sqrt {x}}{x^{2}}}}
7)
h
(
x
)
=
x
2
+
ln
x
x
4
{\displaystyle h(x)={\frac {x^{2}+\ln x}{\sqrt[{4}]{x}}}}
h
′
(
x
)
=
(
x
2
+
ln
x
x
4
)
′
{\displaystyle h'(x)=\left({\frac {x^{2}+\ln x}{\sqrt[{4}]{x}}}\right)'}
=
(
x
2
+
ln
x
)
′
x
4
−
(
x
2
+
ln
x
)
(
x
4
)
′
(
x
4
)
2
{\displaystyle ={\frac {\left(x^{2}+\ln x\right)'{\sqrt[{4}]{x}}-(x^{2}+\ln x)\left({\sqrt[{4}]{x}}\right)'}{({\sqrt[{4}]{x}})^{2}}}}
=
(
2
x
+
1
x
)
x
4
−
(
x
2
+
ln
x
)
1
4
x
3
4
(
x
4
)
2
{\displaystyle ={\frac {(2x+{\frac {1}{x}}){\sqrt[{4}]{x}}-(x^{2}+\ln x){\frac {1}{4{\sqrt[{4}]{x^{3}}}}}}{({\sqrt[{4}]{x}})^{2}}}}
=
(
2
x
+
1
x
)
x
4
−
(
x
2
+
ln
x
)
1
4
x
3
4
x
.
{\displaystyle ={\frac {(2x+{\frac {1}{x}}){\sqrt[{4}]{x}}-(x^{2}+\ln x){\frac {1}{4{\sqrt[{4}]{x^{3}}}}}}{\sqrt {x}}}.}
9)
h
(
x
)
=
x
2
ln
x
+
ln
x
(
x
−
1
)
ln
x
{\displaystyle h(x)={\frac {x^{2}\ln x+\ln x}{(x-1)\ln x}}}
10)
h
(
x
)
=
2
−
2
+
π
1
+
x
2
+
x
3
{\displaystyle h(x)={\frac {2-{\sqrt {2}}+\pi }{1+x^{2}+x^{3}}}}
Composició de funcions [ edit ] Exercicis per aplicar la composició de funcions directament conegut com regla de la cadena .
f
(
x
)
=
{\displaystyle f(x)=}
f
′
(
x
)
=
{\displaystyle f\,'(x)=}
g
(
x
)
=
{\displaystyle g(x)=}
g
′
(
x
)
=
{\displaystyle g\,'(x)=}
(
f
(
g
(
x
)
)
)
′
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
{\displaystyle {\Big (}f{\big (}g(x){\big )}{\Big )}'=f\,'{\big (}g(x){\big )}\cdot g\,'(x)}
Exemples
2)
h
(
x
)
=
(
3
x
2
−
4
)
2
{\displaystyle h(x)=(3x^{2}-4)^{2}}
...
3)
h
(
x
)
=
1
x
2
−
x
{\displaystyle h(x)={\frac {1}{x^{2}-x}}}
...
4)
h
(
x
)
=
ln
(
x
+
1
)
{\displaystyle h(x)=\ln(x+1)}
...
5)
h
(
x
)
=
x
+
2
3
{\displaystyle h(x)={\sqrt[{3}]{x+2}}}
...
6)
h
(
x
)
=
e
x
5
{\displaystyle h(x)={\sqrt[{5}]{e^{x}}}}
...
7)
h
(
x
)
=
ln
(
x
)
{\displaystyle h(x)=\ln({\sqrt {x}})}
...
8)
h
(
x
)
=
1
(
x
+
1
)
2
{\displaystyle h(x)={\frac {1}{(x+1)^{2}}}}
...
9)
h
(
x
)
=
ln
x
+
1
{\displaystyle h(x)={\sqrt {\ln x+1}}}
...
Autoservei d'exercicis [ edit ] A continuació es donen dues taules per servir-nos els nostres propis exercicis que volem practicar, recordant que hi ha eines online per resoldre'ls.
Funcions
f
(
x
)
{\displaystyle f(x)}
g
(
x
)
{\displaystyle g(x)}
f
(
x
)
{\displaystyle f(x)}
P
(
x
)
{\displaystyle P(x)}
x
n
{\displaystyle {\sqrt[{n}]{x}}}
a
x
{\displaystyle a^{x}}
log
a
x
{\displaystyle \log _{a}{x}}
g
(
x
)
{\displaystyle g(x)}
P
(
x
)
{\displaystyle P(x)}
1,1
1,2
1,3
1,4
x
n
{\displaystyle {\sqrt[{n}]{x}}}
2,1
2,2
2,3
2,4
a
x
{\displaystyle a^{x}}
3,1
3,2
3,3
3,4
log
a
x
{\displaystyle \log _{a}{x}}
4,1
4,2
4,3
4,4
Desprès de decidir les funcions corresponents a f i g heu de escollir una de les operacions següents i derivar-la per practicar totes les possibles derivades existents, de fet es poden ajuntar moltes funcions i després derivar-les, però millor a poc a poc.
1.
f
(
x
)
+
g
(
x
)
{\displaystyle f(x)+g(x)}
2.
f
(
x
)
⋅
g
(
x
)
{\displaystyle f(x)\cdot g(x)}
3.
f
(
x
)
g
(
x
)
{\displaystyle {\frac {f(x)}{g(x)}}}
4.
f
∘
g
(
x
)
=
f
(
g
(
x
)
)
{\displaystyle f\circ g(x)=f{\big (}g(x){\big )}}
Exemple:
1.- A la primera taula agafen el requadre 2,3 per tant pot ser
f
(
x
)
=
x
3
{\displaystyle f(x)={\sqrt[{3}]{x}}}
g
(
x
)
=
e
x
.
{\displaystyle g(x)=e^{x}.}
Per tant si a la segona taula escollim el número 3, la nostra derivada a fer és:
d
d
x
(
x
3
e
x
)
=
(
x
3
e
x
)
′
.
{\displaystyle {\frac {d}{dx}}\left({\frac {\sqrt[{3}]{x}}{e^{x}}}\right)=\left({\frac {\sqrt[{3}]{x}}{e^{x}}}\right)'.}
2.- A la primera taula agafen el requadre 3,4 per tant pot ser
f
(
x
)
=
2
x
{\displaystyle f(x)=2^{x}}
g
(
x
)
=
log
5
x
.
{\displaystyle g(x)=\log _{5}x.}
Per tant si a la segona taula escollim el número 4, la nostra derivada a fer és:
d
d
x
(
2
log
5
x
)
=
(
2
log
5
x
)
′
.
{\displaystyle {\frac {d}{dx}}\left(2^{\log _{5}x}\right)=\left(2^{\log _{5}x}\right)'.}
Composicions de funcions successives [ edit ] Derivada de la funció potencial-exponencial [ edit ] Notes i referències [ edit ] 
![{\displaystyle h(x)={\frac {x^{2}+\ln x}{\sqrt[{4}]{x}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23e4f4f440981b5d97a410b5bd6537868db1355c)
![{\displaystyle h'(x)=\left({\frac {x^{2}+\ln x}{\sqrt[{4}]{x}}}\right)'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65a1ba88b0b676b461fa44dcbebb6f8c55a05c9a)
![{\displaystyle ={\frac {\left(x^{2}+\ln x\right)'{\sqrt[{4}]{x}}-(x^{2}+\ln x)\left({\sqrt[{4}]{x}}\right)'}{({\sqrt[{4}]{x}})^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/251de6c93ce27dbe2db2d8906af39e557bbac112)
![{\displaystyle ={\frac {(2x+{\frac {1}{x}}){\sqrt[{4}]{x}}-(x^{2}+\ln x){\frac {1}{4{\sqrt[{4}]{x^{3}}}}}}{({\sqrt[{4}]{x}})^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa1763e416c8019a96a3b0a97f6c591f8cc398e3)
![{\displaystyle ={\frac {(2x+{\frac {1}{x}}){\sqrt[{4}]{x}}-(x^{2}+\ln x){\frac {1}{4{\sqrt[{4}]{x^{3}}}}}}{\sqrt {x}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02f5b77f235fbff8d36d7f31edbe2bcb44c95165)
![{\displaystyle h(x)={\sqrt[{3}]{x+2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dcc80af183aa486bbd5d01e27be7fa8a9bbe0b0)
![{\displaystyle h(x)={\sqrt[{5}]{e^{x}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/366a45e10a9f300a1fa06b4ed5dcad8e2ba3b236)
![{\displaystyle {\sqrt[{n}]{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b3ba2638d05cd9ed8dafae7e34986399e48ea99)
![{\displaystyle f(x)={\sqrt[{3}]{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b1d312e15e009bd6c705519b2b1b7400cf1c368)
![{\displaystyle {\frac {d}{dx}}\left({\frac {\sqrt[{3}]{x}}{e^{x}}}\right)=\left({\frac {\sqrt[{3}]{x}}{e^{x}}}\right)'.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4eee5f560e6458c38d5d81e7d3d09494657ee75)
