A partir de la definició de la derivada d'una funció f (x ):
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{f(x+h)-f(x) \over h}}
Per tant si f (x ) = sin(x )
f
′
(
x
)
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
(
x
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\sin(x+h)-\sin(x) \over h}}
A partir de la identitat trigonomètrica
sin
(
A
+
B
)
=
sin
(
A
)
cos
(
B
)
+
cos
(
A
)
sin
(
B
)
{\displaystyle \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)}
f
′
(
x
)
=
lim
h
→
0
sin
(
x
)
cos
(
h
)
+
cos
(
x
)
sin
(
h
)
−
sin
(
x
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x) \over h}}
Agrupant els termes en cos(x ) i sin(x ), la derivada esdevé
f
′
(
x
)
=
lim
h
→
0
cos
(
x
)
sin
(
h
)
−
sin
(
x
)
(
1
−
cos
(
h
)
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\cos(x)\sin(h)-\sin(x)(1-\cos(h)) \over h}}
Reordenant els termes i el límit s'obté
f
′
(
x
)
=
lim
h
→
0
cos
(
x
)
sin
(
h
)
h
−
lim
h
→
0
sin
(
x
)
(
1
−
cos
(
h
)
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\cos(x)\sin(h) \over h}-\lim _{h\to 0}{\sin(x)(1-\cos(h)) \over h}}
Ara com que sin(x ) i cos(x ) no varien en variar h , es poden treure fora del límit per a obtenir
f
′
(
x
)
=
c
o
s
(
x
)
lim
h
→
0
sin
(
h
)
h
−
sin
(
x
)
lim
h
→
0
(
1
−
cos
(
h
)
)
h
{\displaystyle f'(x)=cos(x)\lim _{h\to 0}{\sin(h) \over h}-\sin(x)\lim _{h\to 0}{(1-\cos(h)) \over h}}
El valor dels límits
lim
h
→
0
sin
(
h
)
h
i
lim
h
→
0
(
1
−
cos
(
h
)
)
h
{\displaystyle \lim _{h\to 0}{\sin(h) \over h}\quad {\text{i}}\quad \lim _{h\to 0}{(1-\cos(h)) \over h}}
Són 1 i 0 respectivament . Per tant, si f (x ) = sin(x ),
f
′
(
x
)
=
cos
(
x
)
{\displaystyle f'(x)=\cos(x)\,}
Si f (x ) = cos(x )
f
′
(
x
)
=
lim
h
→
0
cos
(
x
+
h
)
−
cos
(
x
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\cos(x+h)-\cos(x) \over h}}
A partir de la identitat trigonomètrica
cos
(
A
+
B
)
=
cos
(
A
)
cos
(
B
)
−
sin
(
A
)
sin
(
B
)
{\displaystyle \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)}
f
′
(
x
)
=
lim
h
→
0
cos
(
x
)
cos
(
h
)
−
sin
(
x
)
sin
(
h
)
−
cos
(
x
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x) \over h}}
Operant s'obté
f
′
(
x
)
=
lim
h
→
0
cos
(
x
)
(
cos
(
h
)
−
1
)
−
sin
(
x
)
sin
(
h
)
)
h
{\displaystyle f'(x)=\lim _{h\to 0}{\cos(x)(\cos(h)-1)-\sin(x)\sin(h)) \over h}}
Com que sin(x ) i cos(x ) no varien en variar h , es poden treure fora del límit per a obtenir
f
′
(
x
)
=
c
o
s
(
x
)
lim
h
→
0
cos
(
h
)
−
1
h
−
sin
(
x
)
lim
h
→
0
sin
(
h
)
h
{\displaystyle f'(x)=cos(x)\lim _{h\to 0}{\cos(h)-1 \over h}-\sin(x)\lim _{h\to 0}{\sin(h) \over h}}
El valor dels límits
lim
h
→
0
sin
(
h
)
h
i
lim
h
→
0
(
cos
(
h
)
−
1
)
h
{\displaystyle \lim _{h\to 0}{\sin(h) \over h}\quad {\text{i}}\quad \lim _{h\to 0}{(\cos(h)-1) \over h}}
Són 1 i 0 respectivament. Per tant, si f (x ) = cos(x ),
f
′
(
x
)
=
−
sin
(
x
)
{\displaystyle f'(x)=-\sin(x)\,}
modifica
Si
f
(
x
)
=
cot
(
x
)
{\displaystyle f\left(x\right)=\cot \left(x\right)}
cot
(
x
)
=
1
tan
(
x
)
{\displaystyle \cot \left(x\right)={\frac {1}{\tan \left(x\right)}}}
Regla de la raó inversa d'una funció resulta:
D
1
f
(
x
)
=
−
f
′
(
x
)
f
(
x
)
2
{\displaystyle D{1 \over f(x)}=-{f'(x) \over f(x)^{2}}}
f
′
(
x
)
=
−
sec
2
(
x
)
tan
2
(
x
)
=
−
cos
2
(
x
)
sin
2
(
x
)
1
cos
2
(
x
)
=
−
1
sin
2
(
x
)
=
−
csc
2
(
x
)
{\displaystyle {\begin{aligned}f^{'}\left(x\right)&=-{\frac {\sec ^{2}\left(x\right)}{\tan ^{2}\left(x\right)}}\\&=-{\frac {\cos ^{2}\left(x\right)}{\sin ^{2}\left(x\right)}}{\frac {1}{\cos ^{2}\left(x\right)}}\\&=-{\frac {1}{\sin ^{2}\left(x\right)}}\\&=-\csc ^{2}\left(x\right)\end{aligned}}}
Si
f
(
x
)
=
sec
(
x
)
{\displaystyle f\left(x\right)=\sec \left(x\right)}
sec
(
x
)
=
1
cos
(
x
)
{\displaystyle \sec \left(x\right)={\frac {1}{\cos \left(x\right)}}}
Regla de la raó inversa d'una funció resulta:
D
1
f
(
x
)
=
−
f
′
(
x
)
f
(
x
)
2
{\displaystyle D{1 \over f(x)}=-{f'(x) \over f(x)^{2}}}
f
′
(
x
)
=
−
−
sin
(
x
)
cos
2
(
x
)
=
1
cos
(
x
)
sin
(
x
)
cos
(
x
)
=
sec
(
x
)
tan
(
x
)
{\displaystyle {\begin{aligned}f^{'}\left(x\right)&=-{\frac {-\sin \left(x\right)}{\cos ^{2}\left(x\right)}}\\&={\frac {1}{\cos \left(x\right)}}{\frac {\sin \left(x\right)}{\cos \left(x\right)}}\\&=\sec \left(x\right)\tan \left(x\right)\end{aligned}}}
Si
f
(
x
)
=
csc
(
x
)
{\displaystyle f\left(x\right)=\csc \left(x\right)}
csc
(
x
)
=
1
sin
(
x
)
{\displaystyle \csc \left(x\right)={\frac {1}{\sin \left(x\right)}}}
Regla de la raó inversa d'una funció resulta:
D
1
f
(
x
)
=
−
f
′
(
x
)
f
(
x
)
2
{\displaystyle D{1 \over f(x)}=-{f'(x) \over f(x)^{2}}}
f
′
(
x
)
=
−
cos
(
x
)
sin
2
(
x
)
=
−
1
sin
(
x
)
cos
(
x
)
sin
(
x
)
=
−
csc
(
x
)
cot
(
x
)
{\displaystyle {\begin{aligned}f^{'}\left(x\right)&=-{\frac {\cos \left(x\right)}{\sin ^{2}\left(x\right)}}\\&=-{\frac {1}{\sin \left(x\right)}}{\frac {\cos \left(x\right)}{\sin \left(x\right)}}\\&=-\csc \left(x\right)\cot \left(x\right)\end{aligned}}}
modifica
![{\displaystyle {\frac {d}{dx}}f(x)=f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c88a63b6dca6feb28c3fbfa5db1925eb400bfd4)
![{\displaystyle f'(x)={\frac {\cos(x)\cos(x)-\sin(x)[-\sin(x)]}{\cos ^{2}(x)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8d414bbb87f90f8da3356dd149e56a44c5a0d93)
