La derivata della funzione composta è il prodotto tra la derivata della funzione esterna, avente come argomento la funzione interna, per la derivata della funzione interna:
D
[
f
(
g
(
x
)
)
]
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
.
{\displaystyle \operatorname {D} [f(g(x))]=f'(g(x))\cdot g'(x).}
Le notazioni
D
[
f
(
x
)
]
{\displaystyle \operatorname {D} [f(x)]}
f
′
(
x
)
{\displaystyle f'(x)}
La formula è valida anche per funzioni di più variabili reali e per funzioni vettoriali . Il teorema di derivazione delle funzioni composte afferma che se:
x
(
t
)
=
(
x
1
(
t
)
,
x
2
(
t
)
,
…
,
x
n
(
t
)
)
,
t
∈
R
{\displaystyle \mathbf {x} (t)=(x_{1}(t),x_{2}(t),\dots ,x_{n}(t)),\quad t\in \mathbb {R} }
è un vettore di
R
n
{\displaystyle \mathbb {R} ^{n}}
x
′
(
t
)
=
(
x
1
′
(
t
)
,
x
2
′
(
t
)
,
…
,
x
n
′
(
t
)
)
{\displaystyle \mathbf {x} '(t)=(x'_{1}(t),x'_{2}(t),\dots ,x'_{n}(t))}
e se
f
{\displaystyle f}
funzione differenziabile in
x
(
t
)
{\displaystyle \mathbf {x} (t)}
F
(
t
)
=
f
(
x
(
t
)
)
{\displaystyle F(t)=f(\mathbf {x} (t))}
è differenziabile nella variabile
t
{\displaystyle t}
F
′
(
t
)
=
∑
i
=
1
n
∂
f
(
x
(
t
)
)
∂
x
i
x
i
′
(
t
)
=
⟨
∇
F
(
t
)
,
x
′
(
t
)
⟩
=
⟨
∇
f
(
x
(
t
)
)
,
x
′
(
t
)
⟩
{\displaystyle F'(t)=\sum _{i=1}^{n}{\frac {\partial f(\mathbf {x} (t))}{\partial x_{i}}}x'_{i}(t)=\langle {\mathbf {\nabla } F(t)},{\mathbf {x} '(t)}\rangle =\langle {\mathbf {\nabla } f(\mathbf {x} (t))},{\mathbf {\mathbf {x} } '(t)}\rangle }
dove
∇
f
{\displaystyle \nabla f}
gradiente di
f
{\displaystyle f}
⟨
⋅
,
⋅
⟩
{\displaystyle \langle \cdot ,\cdot \rangle }
prodotto scalare euclideo .
Ad esempio, se
f
{\displaystyle f}
(
g
,
h
)
{\displaystyle (g,h)}
f
(
g
(
t
)
,
h
(
t
)
)
{\displaystyle f(g(t),h(t))}
d
f
d
t
=
∂
f
∂
g
d
g
d
t
+
∂
f
∂
h
d
h
d
t
.
{\displaystyle {df \over dt}={\partial f \over \partial g}{dg \over dt}+{\partial f \over \partial h}{dh \over dt}.}
Inoltre, se
f
{\displaystyle \mathbf {f} }
g
{\displaystyle \mathbf {g} }
J
[
(
f
∘
g
)
(
x
)
]
=
J
[
f
(
g
(
x
)
)
]
⋅
J
[
g
(
x
)
]
,
{\displaystyle J[(\mathbf {f} \circ \mathbf {g} )(x)]=J[\mathbf {f} (\mathbf {g} (x))]\cdot J[\mathbf {g} (x)],}
dove
⋅
{\displaystyle \cdot }
moltiplicazione di matrici e
J
[
f
(
x
)
]
{\displaystyle J[\mathbf {f} (x)]}
matrice jacobiana di
f
{\displaystyle \mathbf {f} }
Sia, per non appesantire la notazione,
Δ
g
:=
g
(
x
+
h
)
−
g
(
x
)
{\displaystyle \Delta g:=g(x+h)-g(x)}
g
(
x
+
h
)
=
g
(
x
)
+
Δ
g
{\displaystyle g(x+h)=g(x)+\Delta g}
ω
(
Δ
g
)
=
{
f
(
g
(
x
)
+
Δ
g
)
−
f
(
g
(
x
)
)
Δ
g
−
f
′
(
g
(
x
)
)
,
se
Δ
g
≠
0
,
0
,
se
Δ
g
=
0.
{\displaystyle \omega (\Delta g)={\begin{cases}{\frac {f(g(x)+\Delta g)-f(g(x))}{\Delta g}}-f'(g(x)),&{\text{se }}\Delta g\neq 0,\\0,&{\text{se }}\Delta g=0.\end{cases}}}
È dunque
f
(
g
(
x
)
+
Δ
g
)
−
f
(
g
(
x
)
)
=
f
′
(
g
(
x
)
)
⋅
Δ
g
+
ω
(
Δ
g
)
⋅
Δ
g
.
{\displaystyle f(g(x)+\Delta g)-f(g(x))=f'(g(x))\cdot \Delta g+\omega (\Delta g)\cdot \Delta g.}
Inoltre, per l'ipotesi di derivabilità di
f
{\displaystyle f}
lim
Δ
g
→
0
ω
(
Δ
g
)
=
0.
{\displaystyle \lim _{\Delta g\to 0}\omega (\Delta g)=0.}
Esaminiamo ora il rapporto incrementale di
f
(
g
(
x
)
)
{\displaystyle f(g(x))}
D
[
f
(
g
(
x
)
)
]
=
lim
h
→
0
f
(
g
(
x
+
h
)
)
−
f
(
g
(
x
)
)
h
=
lim
h
→
0
f
(
g
(
x
)
+
Δ
g
)
−
f
(
g
(
x
)
)
h
=
lim
h
→
0
f
′
(
g
(
x
)
)
⋅
Δ
g
+
ω
(
Δ
g
)
⋅
Δ
g
h
.
{\displaystyle \operatorname {D} [f(g(x))]=\lim _{h\to 0}{\frac {f(g(x+h))-f(g(x))}{h}}=\lim _{h\to 0}{\frac {f(g(x)+\Delta g)-f(g(x))}{h}}=\lim _{h\to 0}{\frac {f'(g(x))\cdot \Delta g+\omega (\Delta g)\cdot \Delta g}{h}}.}
Spezzando la frazione, abbiamo
lim
h
→
0
f
′
(
g
(
x
)
)
⋅
g
(
x
+
h
)
−
g
(
x
)
h
+
lim
h
→
0
ω
(
Δ
g
)
⋅
g
(
x
+
h
)
−
g
(
x
)
h
.
{\displaystyle \lim _{h\to 0}f'(g(x))\cdot {\frac {g(x+h)-g(x)}{h}}+\lim _{h\to 0}\omega (\Delta g)\cdot {\frac {g(x+h)-g(x)}{h}}.}
E quindi passando al limite
D
[
f
(
g
(
x
)
)
]
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
+
0
⋅
g
′
(
x
)
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
.
{\displaystyle \operatorname {D} [f(g(x))]=f'(g(x))\cdot g'(x)+0\cdot g'(x)=f'(g(x))\cdot g'(x).}
Siano
f
:
A
→
R
{\displaystyle f\colon A\to \mathbb {R} }
g
:
B
→
A
{\displaystyle g\colon B\to A}
A
,
B
⊆
R
{\displaystyle A,B\subseteq \mathbb {R} }
Dalla definizione di derivata si ha
D
[
(
f
∘
g
)
(
x
)
]
=
lim
t
→
x
f
(
g
(
t
)
)
−
f
(
g
(
x
)
)
t
−
x
.
{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{t\to x}{\frac {f(g(t))-f(g(x))}{t-x}}.}
L'idea di fondo è dividere il numeratore del rapporto incrementale per
g
(
t
)
−
g
(
x
)
{\displaystyle g(t)-g(x)}
f
{\displaystyle f}
g
(
x
)
{\displaystyle g(x)}
f
{\displaystyle f}
g
(
x
)
{\displaystyle g(x)}
1
{\displaystyle 1}
g
(
t
)
−
g
(
x
)
{\displaystyle g(t)-g(x)}
D
[
(
f
∘
g
)
(
x
)
]
=
lim
t
→
x
f
(
g
(
t
)
)
−
f
(
g
(
x
)
)
t
−
x
g
(
t
)
−
g
(
x
)
g
(
t
)
−
g
(
x
)
.
{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{t\to x}{\frac {f(g(t))-f(g(x))}{t-x}}{\frac {g(t)-g(x)}{g(t)-g(x)}}.}
Per le proprietà associativa e commutativa del prodotto otteniamo:
D
[
(
f
∘
g
)
(
x
)
]
=
lim
t
→
x
f
(
g
(
t
)
)
−
f
(
g
(
x
)
)
g
(
t
)
−
g
(
x
)
g
(
t
)
−
g
(
x
)
t
−
x
.
{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{t\to x}{\frac {f(g(t))-f(g(x))}{g(t)-g(x)}}{\frac {g(t)-g(x)}{t-x}}.}
Poiché per ipotesi
f
{\displaystyle f}
g
{\displaystyle g}
lim
a
→
b
f
(
a
)
−
f
(
b
)
a
−
b
=
D
[
f
(
b
)
]
{\displaystyle \lim _{a\to b}{f(a)-f(b) \over a-b}=\operatorname {D} [f(b)]}
lim
α
→
β
g
(
α
)
−
g
(
β
)
α
−
β
=
D
[
g
(
β
)
]
{\displaystyle \lim _{\alpha \to \beta }{g(\alpha )-g(\beta ) \over \alpha -\beta }=\operatorname {D} [g(\beta )]}
g
(
t
)
:=
θ
{\displaystyle g(t):=\theta }
D
[
(
f
∘
g
)
(
x
)
]
=
lim
θ
→
g
(
x
)
f
(
θ
)
−
f
(
g
(
x
)
)
θ
−
g
(
x
)
lim
t
→
x
g
(
t
)
−
g
(
x
)
t
−
x
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{\theta \to g(x)}{\frac {f(\theta )-f(g(x))}{\theta -g(x)}}\lim _{t\to x}{\frac {g(t)-g(x)}{t-x}}=f'(g(x))\cdot g'(x)}
Si considerino due funzioni
f
,
g
:
[
a
,
b
]
→
R
{\displaystyle f,g\colon [a,b]\to \mathbb {R} }
H
(
x
)
=
g
(
f
(
x
)
)
,
{\displaystyle H(x)=g\left(f(x)\right),}
f
(
x
+
h
)
−
f
(
x
)
=
f
′
(
x
)
h
+
o
1
(
h
)
;
{\displaystyle f(x+h)-f(x)=f'(x)h+o_{1}(h);}
g
(
y
+
k
)
−
g
(
y
)
=
g
′
(
y
)
k
−
o
2
(
k
)
;
{\displaystyle g(y+k)-g(y)=g'(y)k-o_{2}(k);}
H
(
x
+
h
)
−
H
(
x
)
=
α
(
x
)
h
−
o
3
(
h
)
.
{\displaystyle H(x+h)-H(x)=\alpha (x)h-o_{3}(h).}
A questo punto si passa alla riscrittura di
H
(
x
+
h
)
−
H
(
x
)
{\displaystyle H(x+h)-H(x)}
H
(
x
)
=
g
(
f
(
x
)
)
{\displaystyle H(x)=g\left(f(x)\right)}
H
(
x
+
h
)
−
H
(
x
)
=
g
(
f
(
x
+
h
)
)
−
g
(
f
(
x
)
)
.
{\displaystyle H(x+h)-H(x)=g(f(x+h))-g(f(x)).}
Si ricordi che
f
(
x
+
h
)
=
f
(
x
)
+
f
′
(
x
)
h
+
o
1
(
h
)
,
{\displaystyle f(x+h)=f(x)+f'(x)h+o_{1}(h),}
H
(
x
+
h
)
−
H
(
x
)
=
g
(
f
(
x
)
+
f
′
(
x
)
h
+
o
1
(
h
)
)
−
g
(
f
(
x
)
)
.
{\displaystyle H(x+h)-H(x)=g(f(x)+f'(x)h+o_{1}(h))-g(f(x)).}
Si effettua la sostituzione
f
(
x
)
=
y
{\displaystyle f(x)=y}
k
=
f
′
(
x
)
h
+
o
1
(
h
)
{\displaystyle k=f'(x)h+o_{1}(h)}
H
(
x
+
h
)
−
H
(
x
)
=
g
(
y
+
k
)
−
g
(
y
)
=
g
′
(
y
)
k
−
o
2
(
k
)
=
g
′
(
f
(
x
)
)
⋅
(
f
′
(
x
)
h
+
o
1
(
h
)
)
−
o
2
(
k
)
=
g
′
(
f
(
x
)
)
f
′
(
x
)
h
+
g
′
(
f
(
x
)
)
o
1
(
h
)
−
o
2
(
k
)
.
{\displaystyle H(x+h)-H(x)=g(y+k)-g(y)=g'(y)k-o_{2}(k)=g'(f(x))\cdot (f'(x)h+o_{1}(h))-o_{2}(k)=g'(f(x))f'(x)h+g'(f(x))o_{1}(h)-o_{2}(k).}
Si pone
α
(
x
)
=
g
′
(
f
(
x
)
)
f
′
(
x
)
{\displaystyle \alpha (x)=g'(f(x))f'(x)}
o
3
(
h
)
=
g
′
(
f
(
x
)
)
o
1
(
h
)
−
o
2
(
k
)
,
{\displaystyle o_{3}(h)=g'(f(x))o_{1}(h)-o_{2}(k),}
![{\displaystyle \operatorname {D} [f(g(x))]=f'(g(x))\cdot g'(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1d4c3d6dcf2ce587485526b8980d8c04317ccb2)
![{\displaystyle \operatorname {D} [f(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cb5fb9502986cfffac0723962b4cec63fbfdfd4)
![{\displaystyle J[(\mathbf {f} \circ \mathbf {g} )(x)]=J[\mathbf {f} (\mathbf {g} (x))]\cdot J[\mathbf {g} (x)],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69a3876df6afe754933f3cecf3670ec65a54049d)
![{\displaystyle J[\mathbf {f} (x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cd99236d20948b46534a65bbca0029253e342d0)
![{\displaystyle \operatorname {D} [f(g(x))]=\lim _{h\to 0}{\frac {f(g(x+h))-f(g(x))}{h}}=\lim _{h\to 0}{\frac {f(g(x)+\Delta g)-f(g(x))}{h}}=\lim _{h\to 0}{\frac {f'(g(x))\cdot \Delta g+\omega (\Delta g)\cdot \Delta g}{h}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd17f1e39db08495ede5222ed06056217600d18a)
![{\displaystyle \operatorname {D} [f(g(x))]=f'(g(x))\cdot g'(x)+0\cdot g'(x)=f'(g(x))\cdot g'(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3616edf53d4eaf1bc61af336dc04e0ab51bee3a)
![{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{t\to x}{\frac {f(g(t))-f(g(x))}{t-x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60dbf62c3549c5c1b8ffeebf8fbd1d2fc94018c0)
![{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{t\to x}{\frac {f(g(t))-f(g(x))}{t-x}}{\frac {g(t)-g(x)}{g(t)-g(x)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b62e143f1c435478ecd355beeadc822223cd0f0a)
![{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{t\to x}{\frac {f(g(t))-f(g(x))}{g(t)-g(x)}}{\frac {g(t)-g(x)}{t-x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd9d7ca1135be0171756d8ba19341f49a71efc43)
![{\displaystyle \lim _{a\to b}{f(a)-f(b) \over a-b}=\operatorname {D} [f(b)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9415a628fc591197783cd71c754755251803abc2)
![{\displaystyle \lim _{\alpha \to \beta }{g(\alpha )-g(\beta ) \over \alpha -\beta }=\operatorname {D} [g(\beta )]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b59d5fcbc9d21bf8268f091b0b0f186206dae54f)
![{\displaystyle \operatorname {D} [(f\circ g)(x)]=\lim _{\theta \to g(x)}{\frac {f(\theta )-f(g(x))}{\theta -g(x)}}\lim _{t\to x}{\frac {g(t)-g(x)}{t-x}}=f'(g(x))\cdot g'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1326ad6724e32f3fc95e4abecf68ba6d547ca838)
![{\displaystyle f,g\colon [a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/07c9cdd885f66cd36e34a6418e28444dfc32ed27)
![{\displaystyle \operatorname {D} [(f\circ g\circ h)(x)]=f'(g(h(x)))\cdot g'(h(x))\cdot h'(x),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbc8687279f92ee746f633a6395394056d91fbc9)
![{\displaystyle (f\circ g\circ h)(x)=\log \left[\left({x \over 2}\right)^{2}+3{x \over 2}\right]-3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de820da76c1cbe4f2cbc981260d6234b57583f75)
![{\displaystyle \operatorname {D} [(f\circ g\circ h)(x)]={1 \over ({x \over 2})^{2}+3{x \over 2}}\cdot \left(2{x \over 2}+3\right)\cdot {1 \over 2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/488fb0dffc14e3ac68d1ab0482f94196f4ac42ab)
