Derivada d'una constant
modifica
En el cas de la funció constant la seva gràfica és una recta horitzontal i per tant té pendent zero a tot arreu, aquest resultat també s'obté directament en aplicar la definició de derivada a la funció constant: f (x )=c .
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
c
−
c
h
=
0
{\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {c-c}{h}}=0}
Derivada d'una potència entera
modifica
En cas que
f
(
x
)
=
x
n
{\displaystyle f(x)=x^{n}}
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
(
x
+
h
)
n
−
x
n
h
{\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}}
Aplicant la fórmula del binomi de Newton , agrupant els termes que tenen h elevada a una potència superior a 2 i traient h ² factor comú d'aquests termes, resulta:
f
′
(
x
)
=
lim
h
→
0
(
x
n
+
n
h
x
n
−
1
+
h
2
R
)
−
x
n
h
{\displaystyle f'(x)=\lim _{h\to 0}{\frac {(x^{n}+nhx^{n-1}+h^{2}R)-x^{n}}{h}}}
A partir d'aquí, operant s'obté:
f
′
(
x
)
=
lim
h
→
0
n
x
n
−
1
+
h
R
=
n
x
n
−
1
+
lim
h
→
0
h
R
=
n
x
n
−
1
+
0
{\displaystyle f'(x)=\lim _{h\to 0}nx^{n-1}+hR=nx^{n-1}+\lim _{h\to 0}hR=nx^{n-1}+0}
Derivada d'una potència real
modifica
Pel càlcul de la derivada d'una potència real primer es transforma l'expressió:
x
r
=
e
r
ln
(
x
)
{\displaystyle x^{r}=e^{r\ln \left(x\right)}}
Llavors s'aplica la regla de la cadena :
D
(
f
∘
g
)
=
[
(
D
f
)
∘
g
]
D
g
{\displaystyle D\left(f\circ g\right)=\left[\left(Df\right)\circ g\right]Dg}
Amb
f
=
e
x
g
=
r
ln
(
x
)
{\displaystyle f=e^{x}\quad g=r\ln \left(x\right)}
D'aquí, operant, i tenint en compe la derivada de la funció exponencial (vegeu més endavant) resulta:
D
x
r
=
D
e
r
ln
(
x
)
=
[
(
e
x
)
∘
r
ln
(
x
)
]
r
x
=
r
x
e
r
ln
(
x
)
=
r
x
x
r
=
r
x
r
−
1
{\displaystyle {\begin{aligned}Dx^{r}&=De^{r\ln \left(x\right)}\\&=\left[\left(e^{x}\right)\circ r\ln \left(x\right)\right]{\frac {r}{x}}\\&={\frac {r}{x}}e^{r\ln \left(x\right)}\\&={\frac {r}{x}}x^{r}\\&=rx^{r-1}\end{aligned}}}
Aquesta expressió, és formalment idèntica al cas de la potència entera.
Pel cas particular de
r
=
1
/
2
{\displaystyle r=1/2}
f
(
x
)
=
x
=
x
1
/
2
{\displaystyle {\begin{aligned}f\left(x\right)&={\sqrt {x}}\\&=x^{1/2}\end{aligned}}}
Per tant:
f
′
(
x
)
=
1
2
x
(
1
2
−
1
)
=
1
2
x
(
−
1
2
)
=
1
2
x
1
/
2
=
1
2
x
{\displaystyle {\begin{aligned}{f}'\left(x\right)&={\frac {1}{2}}x^{\left({\frac {1}{2}}-1\right)}\\&={\frac {1}{2}}x^{\left(-{\frac {1}{2}}\right)}\\&={\frac {1}{2x^{1/2}}}\\&={\frac {1}{2{\sqrt {x}}}}\end{aligned}}}
Derivada de la funció logaritme
modifica
Sigui
x
>
0
{\displaystyle x>0}
funció logaritmica com
f
(
x
)
=
log
a
(
x
)
{\displaystyle f\left(x\right)=\log _{a}\left(x\right)}
f
′
(
x
)
=
lim
h
→
0
log
a
(
x
+
h
)
−
log
a
(
x
)
h
=
lim
h
→
0
log
a
(
x
+
h
x
)
1
h
=
lim
h
→
0
log
a
(
1
+
h
x
)
1
h
{\displaystyle {\begin{aligned}{f}'\left(x\right)&={\underset {h\to 0}{\mathop {\lim } }}\,{\frac {\log _{a}\left(x+h\right)-\log _{a}\left(x\right)}{h}}\\&={\underset {h\to 0}{\mathop {\lim } }}\,\log _{a}\left({\frac {x+h}{x}}\right)^{\frac {1}{h}}\\&={\underset {h\to 0}{\mathop {\lim } }}\,\log _{a}\left(1+{\frac {h}{x}}\right)^{\frac {1}{h}}\end{aligned}}}
Aquesta expressió es pot transformar de la següent manera:
f
′
(
x
)
=
lim
h
→
0
log
a
[
(
1
+
1
x
h
)
x
h
]
1
x
=
1
x
lim
h
→
0
log
a
(
1
+
1
x
h
)
x
h
=
1
x
log
a
lim
h
→
0
(
1
+
1
x
h
)
x
h
{\displaystyle {\begin{aligned}{f}'\left(x\right)&={\underset {h\to 0}{\mathop {\lim } }}\,\log _{a}\left[\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}\right]^{\frac {1}{x}}\\&={\frac {1}{x}}{\underset {h\to 0}{\mathop {\lim } }}\,\log _{a}\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}\\&={\frac {1}{x}}\log _{a}{\underset {h\to 0}{\mathop {\lim } }}\,\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}\end{aligned}}}
Però quant
h
{\displaystyle h}
x
/
h
{\displaystyle x/h}
x
>
0
{\displaystyle x>0}
nombre e :
lim
h
→
0
(
1
+
1
x
h
)
x
h
=
lim
y
→
∞
(
1
+
1
y
)
y
=
e
{\displaystyle {\underset {h\to 0}{\mathop {\lim } }}\,\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}={\underset {y\to \infty }{\mathop {\lim } }}\,\left(1+{\frac {1}{y}}\right)^{y}=e}
Per tant la derivada de la funció logaritme és:
f
′
(
x
)
=
1
x
log
a
(
e
)
{\displaystyle {f}'\left(x\right)={\frac {1}{x}}\log _{a}\left(e\right)}
O cosa que és el mateix:
f
′
(
x
)
=
1
ln
(
a
)
1
x
{\displaystyle {f}'\left(x\right)={\frac {1}{\ln \left(a\right)}}{\frac {1}{x}}}
tenint en compte que:
log
a
(
e
)
=
1
ln
(
a
)
{\displaystyle \log _{a}\left(e\right)={\frac {1}{\ln \left(a\right)}}}
Com es pot comprovar plantejant:
a
log
a
(
e
)
ln
(
a
)
=
(
a
log
a
(
e
)
)
ln
(
a
)
=
e
ln
(
a
)
=
a
⇒
log
a
(
e
)
ln
(
a
)
=
1
⇒
log
a
(
e
)
=
1
ln
(
a
)
{\displaystyle {\begin{aligned}a^{\log _{a}\left(e\right)\ln \left(a\right)}&=\left(a^{\log _{a}\left(e\right)}\right)^{\ln \left(a\right)}\\&=e^{\ln \left(a\right)}\\&=a\\\Rightarrow \log _{a}\left(e\right)\ln \left(a\right)&=1\\\Rightarrow \log _{a}\left(e\right)&={\frac {1}{\ln \left(a\right)}}\end{aligned}}}
En el cas particular del logaritme natural:
f
(
x
)
=
ln
(
x
)
⇒
f
′
(
x
)
=
1
x
ln
(
e
)
=
1
x
{\displaystyle f\left(x\right)=\ln \left(x\right)\Rightarrow {f}'\left(x\right)={\frac {1}{x}}\ln \left(e\right)={\frac {1}{x}}}
Si en comptes de la funció
f
(
x
)
=
log
a
(
x
)
{\displaystyle f(x)=\log _{a}(x)}
F
(
x
)
=
log
a
(
f
(
x
)
)
{\displaystyle F(x)=\log _{a}(f(x))}
x
>
0
{\displaystyle x>0}
f
(
x
)
>
0
{\displaystyle f(x)>0}
Podem aplicar la regla de la cadena per calcular
F
′
(
x
)
{\displaystyle F'(x)}
F
′
(
x
)
=
d
d
x
F
(
x
)
=
d
F
(
x
)
d
f
(
x
)
⋅
d
f
(
x
)
d
x
=
1
f
(
x
)
1
ln
a
⋅
f
′
(
x
)
=
f
′
(
x
)
f
(
x
)
ln
a
{\displaystyle F'(x)={\frac {d}{dx}}F(x)={\frac {dF(x)}{df(x)}}\cdot {\frac {df(x)}{dx}}={\frac {1}{f(x)}}{\frac {1}{\ln a}}\cdot f'(x)={\frac {f'(x)}{f(x)\ln a}}}
I, en concret si
F
(
x
)
=
ln
(
f
(
x
)
)
{\displaystyle F(x)=\ln {\big (}f(x){\big )}}
F
′
(
x
)
=
d
d
x
F
(
x
)
=
d
F
(
x
)
d
f
(
x
)
⋅
d
f
(
x
)
d
x
=
1
f
(
x
)
1
ln
e
⋅
f
′
(
x
)
=
f
′
(
x
)
f
(
x
)
{\displaystyle F'(x)={\frac {d}{dx}}F(x)={\frac {dF(x)}{df(x)}}\cdot {\frac {df(x)}{dx}}={\frac {1}{f(x)}}{\frac {1}{\ln e}}\cdot f'(x)={\frac {f'(x)}{f(x)}}}
Derivada de la funció exponencial
modifica
Com que la funció exponencial és la inversa de la funció logaritme, s'aplica la regla de la derivada de la funció inversa :
[
f
−
1
]
′
(
x
)
=
1
f
′
[
f
−
1
(
x
)
]
{\displaystyle \left[f^{-1}\right]^{\prime }\left(x\right)={\frac {1}{{f}'\left[f^{-1}\left(x\right)\right]}}}
Amb:
f
(
x
)
=
log
a
(
x
)
f
′
(
x
)
=
1
x
log
a
(
e
)
f
−
1
(
x
)
=
a
x
{\displaystyle f\left(x\right)=\log _{a}\left(x\right)\quad {f}'\left(x\right)={\frac {1}{x}}\log _{a}\left(e\right)\quad f^{-1}\left(x\right)=a^{x}}
Substituint i operant resulta:
[
a
x
]
′
=
1
[
1
x
log
a
(
e
)
]
∘
[
a
x
]
=
1
1
a
x
log
a
(
e
)
=
a
x
log
a
(
e
)
{\displaystyle {\begin{aligned}\left[a^{x}\right]^{\prime }&={\frac {1}{\left[{\frac {1}{x}}\log _{a}\left(e\right)\right]\circ \left[a^{x}\right]}}\\&={\frac {1}{{\frac {1}{a^{x}}}\log _{a}\left(e\right)}}\\&={\frac {a^{x}}{\log _{a}\left(e\right)}}\end{aligned}}}
O cosa que és el mateix:
[
a
x
]
′
=
ln
(
a
)
a
x
{\displaystyle \left[a^{x}\right]^{\prime }=\ln \left(a\right)a^{x}}
Pel cas particular de què
a
=
e
{\displaystyle a=e}
[
e
x
]
′
=
e
x
ln
(
e
)
=
e
x
1
=
e
x
{\displaystyle \left[e^{x}\right]^{\prime }={\frac {e^{x}}{\ln \left(e\right)}}={\frac {e^{x}}{1}}=e^{x}}
De nou, aplicant la regla de la cadena podem trobar la derivada de la funció
F
(
x
)
=
a
f
(
x
)
{\displaystyle F(x)=a^{f(x)}}
F
′
(
x
)
=
d
d
x
F
(
x
)
=
d
F
(
x
)
d
f
(
x
)
⋅
d
f
(
x
)
d
x
=
a
f
(
x
)
ln
a
⋅
f
′
(
x
)
{\displaystyle F'(x)={\frac {d}{dx}}F(x)={\frac {dF(x)}{df(x)}}\cdot {\frac {df(x)}{dx}}=a^{f(x)}\ln a\cdot f'(x)}
I, en concret si
F
(
x
)
=
e
f
(
x
)
{\displaystyle F(x)=e^{f(x)}}
F
′
(
x
)
=
d
d
x
F
(
x
)
=
d
F
(
x
)
d
f
(
x
)
⋅
d
f
(
x
)
d
x
=
e
f
(
x
)
ln
e
⋅
f
′
(
x
)
=
e
f
(
x
)
f
′
(
x
)
{\displaystyle F'(x)={\frac {d}{dx}}F(x)={\frac {dF(x)}{df(x)}}\cdot {\frac {df(x)}{dx}}=e^{f(x)}\ln e\cdot f'(x)=e^{f(x)}f'(x)}
Derivada de la funció
f
(
x
)
g
(
x
)
{\displaystyle f(x)^{g(x)}}
modifica
Si
f
{\displaystyle f}
g
{\displaystyle g}
f
(
x
)
>
0
{\displaystyle f(x)>0}
e
f
(
x
)
{\displaystyle e^{f(x)}}
ln
f
(
x
)
{\displaystyle \ln f(x)}
F
(
x
)
=
f
(
x
)
g
(
x
)
{\displaystyle F(x)=f(x)^{g(x)}}
Tenint en compte que podem escriure
f
(
x
)
g
(
x
)
{\displaystyle f(x)^{g(x)}}
e
g
(
x
)
⋅
ln
f
(
x
)
{\displaystyle e^{g(x)\cdot \ln f(x)}}
F
′
(
x
)
=
e
g
(
x
)
⋅
ln
f
(
x
)
d
d
x
(
g
(
x
)
⋅
ln
f
(
x
)
)
=
f
(
x
)
g
(
x
)
[
g
′
(
x
)
ln
f
(
x
)
+
g
(
x
)
d
d
x
(
ln
f
(
x
)
)
]
=
f
(
x
)
g
(
x
)
[
g
′
(
x
)
ln
f
(
x
)
+
g
(
x
)
f
′
(
x
)
f
(
x
)
]
{\displaystyle F'(x)=e^{g(x)\cdot \ln f(x)}{\frac {d}{dx}}{\big (}g(x)\cdot \ln f(x){\big )}=f(x)^{g(x)}{\Big [}g'(x)\ln f(x)+g(x){\frac {d}{dx}}{\big (}\ln f(x){\big )}{\Big ]}=f(x)^{g(x)}{\bigg [}g'(x)\ln f(x)+g(x){\frac {f'(x)}{f(x)}}{\bigg ]}}
Veiem que efectivament aquesta derivada ens condueix a:
d
d
x
(
x
r
)
=
x
r
[
r
′
ln
x
+
r
x
′
x
]
=
x
r
[
0
+
r
1
x
]
=
r
x
r
x
=
r
x
r
−
1
{\displaystyle {\frac {d}{dx}}(x^{r})=x^{r}{\bigg [}r'\ln x+r{\frac {x'}{x}}{\bigg ]}=x^{r}{\bigg [}0+r{\frac {1}{x}}{\bigg ]}=r{\frac {x^{r}}{x}}=rx^{r-1}}
d
d
x
(
a
x
)
=
a
x
[
x
′
ln
a
+
x
a
′
a
]
=
a
x
[
1
ln
a
+
0
]
=
a
x
ln
a
{\displaystyle {\frac {d}{dx}}(a^{x})=a^{x}{\bigg [}x'\ln a+x{\frac {a'}{a}}{\bigg ]}=a^{x}{\bigg [}1\ln a+0{\bigg ]}=a^{x}\ln a}
Cal notar que la primera expressió només està definida quan
x
>
0
{\displaystyle x>0}
a
>
0
{\displaystyle a>0}
r
{\displaystyle r}
(
−
1
)
r
{\displaystyle (-1)^{r}}
x
<
0
{\displaystyle x<0}
x
r
=
(
−
1
)
r
(
−
x
)
r
{\displaystyle x^{r}=(-1)^{r}(-x)^{r}}
d
d
x
(
x
r
)
=
(
−
1
)
r
d
d
x
(
−
x
)
r
=
(
−
1
)
r
(
−
x
)
r
[
r
′
ln
(
−
x
)
+
r
(
−
x
)
′
−
x
]
=
x
r
[
0
+
r
1
x
]
=
r
x
r
x
=
r
x
r
−
1
{\displaystyle {\frac {d}{dx}}(x^{r})=(-1)^{r}{\frac {d}{dx}}(-x)^{r}=(-1)^{r}(-x)^{r}{\bigg [}r'\ln(-x)+r{\frac {(-x)'}{-x}}{\bigg ]}=x^{r}{\bigg [}0+r{\frac {1}{x}}{\bigg ]}=r{\frac {x^{r}}{x}}=rx^{r-1}}
Generalitzant així la equació per a tot valor de
x
{\displaystyle x}
Les altres expressions es poden trobar de manera similar o aplicant la derivada de la funció inversa.
Derivada de les funcions trigonomètriques
modifica
Les derivades de les funcions sinus i cosinus es troben a partir de la definició de derivada, aplicant les identitats trigonomètriques de la suma de raons trigonomètriques
sin
(
α
+
β
)
=
sin
α
cos
β
+
cos
α
sin
β
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \,}
cos
(
α
+
β
)
=
cos
(
β
)
cos
(
α
)
−
sin
(
β
)
sin
(
α
)
{\displaystyle \cos(\alpha +\beta )=\cos \left(\beta \right)\cos \left(\alpha \right)-\sin \left(\beta \right)\sin(\alpha )}
i les identitats trigonomètriques
lim
θ
→
0
sin
θ
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
lim
θ
→
0
1
−
cos
θ
θ
=
0
{\displaystyle \lim _{\theta \to 0}{\frac {1-\cos \theta }{\theta }}=0\,}
Un cop s'han trobat les derivades del sinus i del cosinus la derivada de la tangent es calcula aplicant la regla del quocient a la identitat trigonomètrica:
tg
(
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle \operatorname {tg} (x)={\frac {\sin(x)}{\cos(x)}}}
A partir d'aqui es troben les derivades de les funcions cotangent, secant i cosecant aplicant la Regla de la raó inversa d'una funció a les identitats:
cotg
(
x
)
=
1
tg
(
x
)
sec
(
x
)
=
1
cos
(
x
)
i
cosec
(
x
)
=
1
sin
(
x
)
{\displaystyle \operatorname {cotg} (x)={\frac {1}{\operatorname {tg} (x)}}\quad \sec(x)={\frac {1}{\cos \left(x\right)}}\quad i\quad \operatorname {cosec} (x)={\frac {1}{\sin(x)}}}
Els detalls de tot el procés es troben a l'article Derivació de les funcions trigonomètriques
Derivada de les funcions inverses de les funcions trigonomètriques
modifica
La derivada de les inverses de les funcions trigonomètriques es calculen aplicant la regla de la funció inversa a cada una de les funcions trigonomètriques i simplificant el resultat.
Derivada de les funcions hiperbòliques
modifica
Les derivades de les funcions hiperbòliques s'obtenen a partir de les seves definicions emprant la derivada de la funció
F
(
x
)
=
e
f
(
x
)
{\displaystyle F(x)=e^{f(x)}}
(
sinh
x
)
′
{\displaystyle (\sinh x)'}
modifica
d
d
x
sinh
x
=
d
d
x
(
e
x
−
e
−
x
2
)
=
e
x
+
e
−
x
2
=
cosh
x
{\displaystyle {\frac {d}{dx}}\sinh x={\frac {d}{dx}}{\bigg (}{\frac {e^{x}-e^{-x}}{2}}{\bigg )}={\frac {e^{x}+e^{-x}}{2}}=\cosh x}
(
cosh
x
)
′
{\displaystyle (\cosh x)'}
modifica
d
d
x
cosh
x
=
d
d
x
(
e
x
+
e
−
x
2
)
=
e
x
−
e
−
x
2
=
sinh
x
{\displaystyle {\frac {d}{dx}}\cosh x={\frac {d}{dx}}{\bigg (}{\frac {e^{x}+e^{-x}}{2}}{\bigg )}={\frac {e^{x}-e^{-x}}{2}}=\sinh x}
(
tanh
x
)
′
{\displaystyle (\tanh x)'}
modifica
d
d
x
tanh
x
=
d
d
x
(
sinh
x
cosh
x
)
=
(
sinh
x
)
′
(
cosh
x
)
−
(
sinh
x
)
(
cosh
x
)
′
cosh
2
x
=
cosh
2
x
−
sinh
2
x
cosh
2
x
=
1
−
sinh
2
x
cosh
2
x
=
1
−
tanh
2
x
{\displaystyle {\frac {d}{dx}}\tanh x={\frac {d}{dx}}{\bigg (}{\frac {\sinh x}{\cosh x}}{\bigg )}={\frac {(\sinh x)'(\cosh x)-(\sinh x)(\cosh x)'}{\cosh ^{2}x}}={\frac {\cosh ^{2}x-\sinh ^{2}x}{\cosh ^{2}x}}=1-{\frac {\sinh ^{2}x}{\cosh ^{2}x}}=1-\tanh ^{2}x}
O, utilitzant la relació
cosh
2
x
−
sinh
2
x
=
1
{\displaystyle \cosh ^{2}x-\sinh ^{2}x=1}
d
d
x
tanh
x
=
cosh
2
x
−
sinh
2
x
cosh
2
x
=
1
cosh
2
x
=
sech
2
x
{\displaystyle {\frac {d}{dx}}\tanh x={\frac {\cosh ^{2}x-\sinh ^{2}x}{\cosh ^{2}x}}={\frac {1}{\cosh ^{2}x}}={\hbox{sech}}^{2}x}
Per les funcions hiperbòliques inverses fem servir la regla de la funció inversa .
(
arcsinh
x
)
′
{\displaystyle ({\hbox{arcsinh}}x)'}
modifica
Denotem
y
=
arcsinh
(
x
)
⟹
x
=
sinh
y
{\displaystyle y={\hbox{arcsinh}}(x)\Longrightarrow x=\sinh y}
d
y
d
x
=
1
d
x
/
d
y
{\displaystyle {\frac {dy}{dx}}={\frac {1}{dx/dy}}}
d
d
x
arcsinh
(
x
)
=
1
d
/
d
y
(
sinh
y
)
=
1
cosh
y
=
1
cosh
(
arcsinh
(
x
)
)
{\displaystyle {\frac {d}{dx}}{\hbox{arcsinh}}(x)={\frac {1}{d/dy(\sinh y)}}={\frac {1}{\cosh y}}={\frac {1}{\cosh({\hbox{arcsinh}}(x))}}}
Com que
cosh
x
=
1
+
sinh
2
x
{\displaystyle \cosh x={\sqrt {1+\sinh ^{2}x}}}
d
d
x
arcsinh
(
x
)
=
1
cosh
(
arcsinh
(
x
)
)
=
1
1
+
sinh
2
(
(
arcsinh
(
x
)
)
)
=
1
1
+
x
2
{\displaystyle {\frac {d}{dx}}{\hbox{arcsinh}}(x)={\frac {1}{\cosh({\hbox{arcsinh}}(x))}}={\frac {1}{\sqrt {1+\sinh ^{2}(({\hbox{arcsinh}}(x)))}}}={\frac {1}{\sqrt {1+x^{2}}}}}
(
arccosh
x
)
′
{\displaystyle ({\hbox{arccosh}}x)'}
modifica
Sigui
y
=
arccosh
(
x
)
⟹
x
=
cosh
y
{\displaystyle y={\hbox{arccosh}}(x)\Longrightarrow x=\cosh y}
d
y
d
x
=
1
d
x
/
d
y
{\displaystyle {\frac {dy}{dx}}={\frac {1}{dx/dy}}}
d
d
x
arccosh
(
x
)
=
1
d
/
d
y
(
cosh
y
)
=
1
sinh
y
=
1
sinh
(
arccosh
(
x
)
)
{\displaystyle {\frac {d}{dx}}{\hbox{arccosh}}(x)={\frac {1}{d/dy(\cosh y)}}={\frac {1}{\sinh y}}={\frac {1}{\sinh({\hbox{arccosh}}(x))}}}
Com que
sinh
x
=
cosh
2
(
x
)
−
1
{\displaystyle \sinh x={\sqrt {\cosh ^{2}(x)-1}}}
d
d
x
arccosh
(
x
)
=
1
sinh
(
arccosh
(
x
)
)
=
1
cosh
2
(
(
arccosh
(
x
)
)
)
−
1
=
1
x
2
−
1
{\displaystyle {\frac {d}{dx}}{\hbox{arccosh}}(x)={\frac {1}{\sinh({\hbox{arccosh}}(x))}}={\frac {1}{\sqrt {\cosh ^{2}(({\hbox{arccosh}}(x)))-1}}}={\frac {1}{\sqrt {x^{2}-1}}}}
(
arctanh
x
)
′
{\displaystyle ({\hbox{arctanh}}x)'}
modifica
Sigui
y
=
arctanh
(
x
)
⟹
x
=
tanh
y
{\displaystyle y={\hbox{arctanh}}(x)\Longrightarrow x=\tanh y}
d
y
d
x
=
1
d
x
/
d
y
{\displaystyle {\frac {dy}{dx}}={\frac {1}{dx/dy}}}
d
d
x
arctanh
(
x
)
=
1
d
/
d
y
(
tanh
y
)
=
1
1
−
tanh
2
y
=
1
1
−
tanh
2
(
arctanh
(
x
)
)
=
1
1
−
x
2
{\displaystyle {\frac {d}{dx}}{\hbox{arctanh}}(x)={\frac {1}{d/dy(\tanh y)}}={\frac {1}{1-\tanh ^{2}y}}={\frac {1}{1-\tanh ^{2}({\hbox{arctanh}}(x))}}={\frac {1}{1-x^{2}}}}
![{\displaystyle D\left(f\circ g\right)=\left[\left(Df\right)\circ g\right]Dg}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d661950c223f354144f871b57ac253f4850f12ad)
![{\displaystyle {\begin{aligned}Dx^{r}&=De^{r\ln \left(x\right)}\\&=\left[\left(e^{x}\right)\circ r\ln \left(x\right)\right]{\frac {r}{x}}\\&={\frac {r}{x}}e^{r\ln \left(x\right)}\\&={\frac {r}{x}}x^{r}\\&=rx^{r-1}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b00e4cff9d59486e53bce9c94dcd30b73cba448)
![{\displaystyle {\begin{aligned}{f}'\left(x\right)&={\underset {h\to 0}{\mathop {\lim } }}\,\log _{a}\left[\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}\right]^{\frac {1}{x}}\\&={\frac {1}{x}}{\underset {h\to 0}{\mathop {\lim } }}\,\log _{a}\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}\\&={\frac {1}{x}}\log _{a}{\underset {h\to 0}{\mathop {\lim } }}\,\left(1+{\frac {1}{\frac {x}{h}}}\right)^{\frac {x}{h}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c116cc796ea9f5fc842db110f60f4829d06f4f9)
![{\displaystyle \left[f^{-1}\right]^{\prime }\left(x\right)={\frac {1}{{f}'\left[f^{-1}\left(x\right)\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/138979a3ee9e8137a2e89f7dde6cf1fbbac5a68f)
![{\displaystyle {\begin{aligned}\left[a^{x}\right]^{\prime }&={\frac {1}{\left[{\frac {1}{x}}\log _{a}\left(e\right)\right]\circ \left[a^{x}\right]}}\\&={\frac {1}{{\frac {1}{a^{x}}}\log _{a}\left(e\right)}}\\&={\frac {a^{x}}{\log _{a}\left(e\right)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c382038a878a3e5584db3bf910ae629355cef683)
![{\displaystyle \left[a^{x}\right]^{\prime }=\ln \left(a\right)a^{x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/959ac6e3cbcc8f9c0e202cec1cb120b258f2a45f)
![{\displaystyle \left[e^{x}\right]^{\prime }={\frac {e^{x}}{\ln \left(e\right)}}={\frac {e^{x}}{1}}=e^{x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7f2f2d38773b3385ec8df79329b12f3e7d31e19)
![{\displaystyle F'(x)=e^{g(x)\cdot \ln f(x)}{\frac {d}{dx}}{\big (}g(x)\cdot \ln f(x){\big )}=f(x)^{g(x)}{\Big [}g'(x)\ln f(x)+g(x){\frac {d}{dx}}{\big (}\ln f(x){\big )}{\Big ]}=f(x)^{g(x)}{\bigg [}g'(x)\ln f(x)+g(x){\frac {f'(x)}{f(x)}}{\bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5204b560eb0b5337d1b750c38dbc9121971d6bc5)
![{\displaystyle {\frac {d}{dx}}(x^{r})=x^{r}{\bigg [}r'\ln x+r{\frac {x'}{x}}{\bigg ]}=x^{r}{\bigg [}0+r{\frac {1}{x}}{\bigg ]}=r{\frac {x^{r}}{x}}=rx^{r-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b8ed6d94c4dd24a7cb598a2e2ba30d0ddfd7df2)
![{\displaystyle {\frac {d}{dx}}(a^{x})=a^{x}{\bigg [}x'\ln a+x{\frac {a'}{a}}{\bigg ]}=a^{x}{\bigg [}1\ln a+0{\bigg ]}=a^{x}\ln a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c87a54b5f429d3307f6d4aad3115407db06621c)
![{\displaystyle {\frac {d}{dx}}(x^{r})=(-1)^{r}{\frac {d}{dx}}(-x)^{r}=(-1)^{r}(-x)^{r}{\bigg [}r'\ln(-x)+r{\frac {(-x)'}{-x}}{\bigg ]}=x^{r}{\bigg [}0+r{\frac {1}{x}}{\bigg ]}=r{\frac {x^{r}}{x}}=rx^{r-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b28d345eac2d5955ede6004b283291662ad864f)
