En el procés de càlcul de derivades o diferenciació, es pot obtenir la derivada de qualsevol funció elemental emprant les regles de derivació i la taula de derivades de les funcions base a partir de les quals es construeixen la resta de funcions elementals.
Les derivades d'aquestes funcions base s'obtenen normalment a partir de la definició de derivada, aplicant les propietats de cada funció i amb les tècniques de càlcul de límits .
Funció F : primitiva de f
Funció f : derivada de F
Funcions elementals
f
(
x
)
=
k
{\displaystyle f(x)=k\,}
f
′
(
x
)
=
0
{\displaystyle f'(x)=0\,}
f
(
x
)
=
x
{\displaystyle f(x)=x\,}
f
′
(
x
)
=
1
{\displaystyle f'(x)=1\,}
f
(
x
)
=
x
n
{\displaystyle f(x)=x^{n}\,}
f
′
(
x
)
=
n
x
n
−
1
{\displaystyle f'(x)=nx^{n-1}\,}
f
(
x
)
=
x
{\displaystyle f(x)={\sqrt {x}}\,}
f
′
(
x
)
=
1
2
x
{\displaystyle f'(x)={\frac {1}{2{\sqrt {x}}}}\,}
f
(
x
)
=
e
x
{\displaystyle f(x)=e^{x}\,}
f
′
(
x
)
=
e
x
{\displaystyle f'(x)=e^{x}\,}
f
(
x
)
=
ln
(
x
)
{\displaystyle f(x)=\ln(x)\,}
f
′
(
x
)
=
1
x
{\displaystyle f'(x)={\frac {1}{x}}\,}
f
(
x
)
=
a
x
(amb
a
>
0
)
{\displaystyle f(x)=a^{x}\quad {\text{(amb }}a>0)\,}
f
′
(
x
)
=
a
x
ln
(
a
)
{\displaystyle f'(x)=a^{x}\ln(a)\,}
f
(
x
)
=
log
b
(
x
)
{\displaystyle f(x)=\log _{b}(x)\,}
f
′
(
x
)
=
1
x
ln
(
b
)
{\displaystyle f'(x)={\frac {1}{x\ln(b)}}\,}
f
(
x
)
=
1
x
n
=
(
x
n
)
−
1
=
x
−
n
{\displaystyle f(x)={\frac {1}{x^{n}}}=(x^{n})^{-1}=x^{-n}\,}
f
′
(
x
)
=
−
n
x
−
n
−
1
{\displaystyle f'(x)=-nx^{-n-1}\,}
Funcions trigonomètriques
f
(
x
)
=
sin
(
x
)
{\displaystyle f(x)=\sin(x)\,}
f
′
(
x
)
=
cos
(
x
)
{\displaystyle f'(x)=\cos(x)\,}
f
(
x
)
=
cos
(
x
)
{\displaystyle f(x)=\cos(x)\,}
f
′
(
x
)
=
−
sin
(
x
)
{\displaystyle f'(x)=-\sin(x)\,}
f
(
x
)
=
tg
(
x
)
{\displaystyle f(x)=\operatorname {tg} (x)\,}
f
′
(
x
)
=
sec
2
(
x
)
{\displaystyle f'(x)=\sec ^{2}(x)\,}
f
(
x
)
=
sec
(
x
)
{\displaystyle f(x)=\sec(x)\,}
f
′
(
x
)
=
sec
(
x
)
tg
(
x
)
{\displaystyle f'(x)=\sec(x)\operatorname {tg} (x)\,}
f
(
x
)
=
cosec
(
x
)
{\displaystyle f(x)=\operatorname {cosec} (x)\,}
f
′
(
x
)
=
−
cosec
(
x
)
cotg
(
x
)
{\displaystyle f'(x)=-\operatorname {cosec} (x)\operatorname {cotg} (x)\,}
f
(
x
)
=
cotg
(
x
)
{\displaystyle f(x)=\operatorname {cotg} (x)\,}
f
′
(
x
)
=
−
cosec
2
(
x
)
{\displaystyle f'(x)=-\operatorname {cosec} ^{2}(x)\,}
f
(
x
)
=
arcsin
(
x
)
{\displaystyle f(x)=\arcsin(x)\,}
f
′
(
x
)
=
1
1
−
x
2
{\displaystyle f'(x)={\frac {1}{\sqrt {1-x^{2}}}}\,}
f
(
x
)
=
arccos
(
x
)
{\displaystyle f(x)=\arccos(x)\,}
f
′
(
x
)
=
−
1
1
−
x
2
{\displaystyle f'(x)={\frac {-1}{\sqrt {1-x^{2}}}}\,}
f
(
x
)
=
arctg
(
x
)
{\displaystyle f(x)=\operatorname {arctg} (x)\,}
f
′
(
x
)
=
1
1
+
x
2
{\displaystyle f'(x)={\frac {1}{1+x^{2}}}\,}
Funcions hiperbòliques
f
(
x
)
=
sinh
x
{\displaystyle f(x)=\sinh x\,}
f
′
(
x
)
=
cosh
x
=
e
x
+
e
−
x
2
{\displaystyle f'(x)=\cosh x={\frac {e^{x}+e^{-x}}{2}}\,}
f
(
x
)
=
arsinh
x
{\displaystyle f(x)=\operatorname {arsinh} \,x\,}
f
′
(
x
)
=
1
x
2
+
1
{\displaystyle f'(x)={1 \over {\sqrt {x^{2}+1}}}\,}
f
(
x
)
=
cosh
x
{\displaystyle f(x)=\cosh x\,}
f
′
(
x
)
=
sinh
x
=
e
x
−
e
−
x
2
{\displaystyle f'(x)=\sinh x={\frac {e^{x}-e^{-x}}{2}}\,}
f
(
x
)
=
arcosh
x
{\displaystyle f(x)=\operatorname {arcosh} \,x\,}
f
′
(
x
)
=
1
x
2
−
1
{\displaystyle f'(x)={\frac {1}{\sqrt {x^{2}-1}}}\,}
f
(
x
)
=
tgh
x
{\displaystyle f(x)=\operatorname {tgh} \,x\,}
f
′
(
x
)
=
sech
2
x
{\displaystyle f'(x)={\operatorname {sech} ^{2}\,x}\,}
f
(
x
)
=
artgh
x
{\displaystyle f(x)=\operatorname {artgh} \,x\,}
f
′
(
x
)
=
1
1
−
x
2
{\displaystyle f'(x)={1 \over 1-x^{2}}\,}
f
(
x
)
=
sech
x
{\displaystyle f(x)=\operatorname {sech} \,x\,}
f
′
(
x
)
=
−
tgh
x
sech
x
{\displaystyle f'(x)=-\operatorname {tgh} \,x\,\operatorname {sech} \,x\,}
f
(
x
)
=
arsech
x
{\displaystyle f(x)=\operatorname {arsech} \,x\,}
f
′
(
x
)
=
−
1
x
1
−
x
2
{\displaystyle f'(x)=-{1 \over x{\sqrt {1-x^{2}}}}\,}
f
(
x
)
=
cosech
x
{\displaystyle f(x)=\operatorname {cosech} \,x\,}
f
′
(
x
)
=
−
cotgh
x
cosech
x
{\displaystyle f'(x)=-\,\operatorname {cotgh} \,x\,\operatorname {cosech} \,x\,}
f
(
x
)
=
arcosech
x
{\displaystyle f(x)=\operatorname {arcosech} \,x}
f
′
(
x
)
=
−
1
|
x
|
1
+
x
2
{\displaystyle f'(x)=-{1 \over |x|{\sqrt {1+x^{2}}}}\,}
f
(
x
)
=
cotgh
x
{\displaystyle f(x)=\operatorname {cotgh} \,x\,}
f
′
(
x
)
=
−
cosech
2
x
{\displaystyle f'(x)=-\,\operatorname {cosech} ^{2}\,x\,}
f
(
x
)
=
arcotgh
x
{\displaystyle f(x)=\operatorname {arcotgh} \,x\,}
f
′
(
x
)
=
1
1
−
x
2
{\displaystyle f'(x)={1 \over 1-x^{2}}\,}
Funció Gamma
(
Γ
(
x
)
)
′
=
∫
0
∞
t
x
−
1
e
−
t
ln
t
d
t
{\displaystyle (\Gamma (x))'=\int _{0}^{\infty }t^{x-1}e^{-t}\ln t\,dt}
(
Γ
(
x
)
)
′
=
Γ
(
x
)
(
∑
n
=
1
∞
(
ln
(
1
+
1
n
)
−
1
x
+
n
)
−
1
x
)
=
Γ
(
x
)
ψ
(
x
)
{\displaystyle (\Gamma (x))'=\Gamma (x)\left(\sum _{n=1}^{\infty }\left(\ln \left(1+{\dfrac {1}{n}}\right)-{\dfrac {1}{x+n}}\right)-{\dfrac {1}{x}}\right)=\Gamma (x)\psi (x)}
Funció zeta de Riemann
(
ζ
(
x
)
)
′
=
−
∑
n
=
1
∞
ln
n
n
x
=
−
ln
2
2
x
−
ln
3
3
x
−
ln
4
4
x
−
⋯
{\displaystyle (\zeta (x))'=-\sum _{n=1}^{\infty }{\frac {\ln n}{n^{x}}}=-{\frac {\ln 2}{2^{x}}}-{\frac {\ln 3}{3^{x}}}-{\frac {\ln 4}{4^{x}}}-\cdots \!}
(
ζ
(
x
)
)
′
=
−
∑
p
primer
p
−
x
ln
p
(
1
−
p
−
x
)
2
∏
q
primer
,
q
≠
p
1
1
−
q
−
x
{\displaystyle (\zeta (x))'=-\sum _{p{\text{ primer}}}{\frac {p^{-x}\ln p}{(1-p^{-x})^{2}}}\prod _{q{\text{ primer}},q\neq p}{\frac {1}{1-q^{-x}}}\!}
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}
[ 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}
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}}}
[ 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)}}}
[ 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)}
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.
[ 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
[ 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.
[ 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}}}}
http://www.edicionsupc.cat/virtuals/caplln/ME01007X.htm# [Enllaç no actiu ]
