An illustration of the product rule: the increment of the area is about the seize of the sum of the two products of the side length and the increment of the other side length. For the infinitesimal increment, the product of the two increments is irrelevant.
Let
D
⊆
R
{\displaystyle {}D\subseteq \mathbb {R} }
a
∈
D
{\displaystyle {}a\in D}
f
,
g
:
D
⟶
R
{\displaystyle f,g\colon D\longrightarrow \mathbb {R} }
functions
which are
differentiable
in
a
{\displaystyle {}a}
. Then the following rules for differentiability holds.
The sum
f
+
g
{\displaystyle {}f+g}
a
{\displaystyle {}a}
(
f
+
g
)
′
(
a
)
=
f
′
(
a
)
+
g
′
(
a
)
.
{\displaystyle {}(f+g)'(a)=f'(a)+g'(a)\,.}
The product
f
⋅
g
{\displaystyle {}f\cdot g}
a
{\displaystyle {}a}
(
f
⋅
g
)
′
(
a
)
=
f
′
(
a
)
g
(
a
)
+
f
(
a
)
g
′
(
a
)
.
{\displaystyle {}(f\cdot g)'(a)=f'(a)g(a)+f(a)g'(a)\,.}
For
c
∈
R
{\displaystyle {}c\in \mathbb {R} }
c
f
{\displaystyle {}cf}
a
{\displaystyle {}a}
(
c
f
)
′
(
a
)
=
c
f
′
(
a
)
.
{\displaystyle {}(cf)'(a)=cf'(a)\,.}
If
g
{\displaystyle {}g}
a
{\displaystyle {}a}
1
/
g
{\displaystyle {}1/g}
a
{\displaystyle {}a}
(
1
g
)
′
(
a
)
=
−
g
′
(
a
)
(
g
(
a
)
)
2
.
{\displaystyle {}{\left({\frac {1}{g}}\right)}'(a)={\frac {-g'(a)}{(g(a))^{2}}}\,.}
If
g
{\displaystyle {}g}
a
{\displaystyle {}a}
f
/
g
{\displaystyle {}f/g}
a
{\displaystyle {}a}
(
f
g
)
′
(
a
)
=
f
′
(
a
)
g
(
a
)
−
f
(
a
)
g
′
(
a
)
(
g
(
a
)
)
2
.
{\displaystyle {}{\left({\frac {f}{g}}\right)}'(a)={\frac {f'(a)g(a)-f(a)g'(a)}{(g(a))^{2}}}\,.}
(1). We write
f
{\displaystyle {}f}
g
{\displaystyle {}g}
fact ,
that is
f
(
x
)
=
f
(
a
)
+
s
(
x
−
a
)
+
r
(
x
)
(
x
−
a
)
{\displaystyle {}f(x)=f(a)+s(x-a)+r(x)(x-a)\,}
and
g
(
x
)
=
g
(
a
)
+
s
~
(
x
−
a
)
+
r
~
(
x
)
(
x
−
a
)
.
{\displaystyle {}g(x)=g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a)\,.}
Summing up yields
f
(
x
)
+
g
(
x
)
=
f
(
a
)
+
g
(
a
)
+
(
s
+
s
~
)
(
x
−
a
)
+
(
r
+
r
~
)
(
x
)
(
x
−
a
)
.
{\displaystyle {}f(x)+g(x)=f(a)+g(a)+(s+{\tilde {s}})(x-a)+(r+{\tilde {r}})(x)(x-a)\,.}
Here, the sum
r
+
r
~
{\displaystyle {}r+{\tilde {r}}}
a
{\displaystyle {}a}
0
{\displaystyle {}0}
f
(
x
)
=
f
(
a
)
+
s
(
x
−
a
)
+
r
(
x
)
(
x
−
a
)
{\displaystyle {}f(x)=f(a)+s(x-a)+r(x)(x-a)\,}
and
g
(
x
)
=
g
(
a
)
+
s
~
(
x
−
a
)
+
r
~
(
x
)
(
x
−
a
)
,
{\displaystyle {}g(x)=g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a)\,,}
and multiply both equations. This yields
f
(
x
)
g
(
x
)
=
(
f
(
a
)
+
s
(
x
−
a
)
+
r
(
x
)
(
x
−
a
)
)
(
g
(
a
)
+
s
~
(
x
−
a
)
+
r
~
(
x
)
(
x
−
a
)
)
=
f
(
a
)
g
(
a
)
+
(
s
g
(
a
)
+
s
~
f
(
a
)
)
(
x
−
a
)
+
(
f
(
a
)
r
~
(
x
)
+
g
(
a
)
r
(
x
)
+
s
s
~
(
x
−
a
)
+
s
r
~
(
x
)
(
x
−
a
)
+
s
~
r
(
x
)
(
x
−
a
)
+
r
(
x
)
r
~
(
x
)
(
x
−
a
)
)
(
x
−
a
)
.
{\displaystyle {}{\begin{aligned}f(x)g(x)&=(f(a)+s(x-a)+r(x)(x-a))(g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a))\\&=f(a)g(a)+(sg(a)+{\tilde {s}}f(a))(x-a)\\&\,\,\,\,\,+(f(a){\tilde {r}}(x)+g(a)r(x)+s{\tilde {s}}(x-a)+s{\tilde {r}}(x)(x-a)+{\tilde {s}}r(x)(x-a)+r(x){\tilde {r}}(x)(x-a))(x-a).\end{aligned}}}
Due to
fact
for
limits,
the expression consisting of the last six summands is a continuous function, with value
0
{\displaystyle {}0}
x
=
a
{\displaystyle {}x=a}
0
{\displaystyle {}0}
1
g
(
x
)
−
1
g
(
a
)
x
−
a
=
−
1
g
(
a
)
g
(
x
)
⋅
g
(
x
)
−
g
(
a
)
x
−
a
.
{\displaystyle {}{\frac {{\frac {1}{g(x)}}-{\frac {1}{g(a)}}}{x-a}}={\frac {-1}{g(a)g(x)}}\cdot {\frac {g(x)-g(a)}{x-a}}\,.}
Since
g
{\displaystyle {}g}
a
{\displaystyle {}a}
fact ,
the left-hand factor converges for
x
→
a
{\displaystyle {}x\rightarrow a}
−
1
g
(
a
)
2
{\displaystyle {}-{\frac {1}{g(a)^{2}}}}
g
{\displaystyle {}g}
a
{\displaystyle {}a}
g
′
(
a
)
{\displaystyle {}g'(a)}
◻
{\displaystyle \Box }
sum rule , product rule , quotient rule . The following statement is called chain rule .
Due to
fact ,
one can write
f
(
x
)
=
f
(
a
)
+
f
′
(
a
)
(
x
−
a
)
+
r
(
x
)
(
x
−
a
)
{\displaystyle {}f(x)=f(a)+f'(a)(x-a)+r(x)(x-a)\,}
and
g
(
y
)
=
g
(
b
)
+
g
′
(
b
)
(
y
−
b
)
+
s
(
y
)
(
y
−
b
)
.
{\displaystyle {}g(y)=g(b)+g'(b)(y-b)+s(y)(y-b)\,.}
Therefore,
g
(
f
(
x
)
)
=
g
(
f
(
a
)
)
+
g
′
(
f
(
a
)
)
(
f
(
x
)
−
f
(
a
)
)
+
s
(
f
(
x
)
)
(
f
(
x
)
−
f
(
a
)
)
=
g
(
f
(
a
)
)
+
g
′
(
f
(
a
)
)
(
f
′
(
a
)
(
x
−
a
)
+
r
(
x
)
(
x
−
a
)
)
+
s
(
f
(
x
)
)
(
f
′
(
a
)
(
x
−
a
)
+
r
(
x
)
(
x
−
a
)
)
=
g
(
f
(
a
)
)
+
g
′
(
f
(
a
)
)
f
′
(
a
)
(
x
−
a
)
+
(
g
′
(
f
(
a
)
)
r
(
x
)
+
s
(
f
(
x
)
)
(
f
′
(
a
)
+
r
(
x
)
)
)
(
x
−
a
)
.
{\displaystyle {}{\begin{aligned}g(f(x))&=g(f(a))+g'(f(a))(f(x)-f(a))+s(f(x))(f(x)-f(a))\\&=g(f(a))+g'(f(a)){\left(f'(a)(x-a)+r(x)(x-a)\right)}+s(f(x)){\left(f'(a)(x-a)+r(x)(x-a)\right)}\\&=g(f(a))+g'(f(a))f'(a)(x-a)+{\left(g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\right)}(x-a).\end{aligned}}}
The remainder function
t
(
x
)
:=
g
′
(
f
(
a
)
)
r
(
x
)
+
s
(
f
(
x
)
)
(
f
′
(
a
)
+
r
(
x
)
)
{\displaystyle {}t(x):=g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\,}
is continuous in
a
{\displaystyle {}a}
0
{\displaystyle {}0}
◻
{\displaystyle \Box }
An illustration for the derivative of the inverse function. The graph of the inverse function is the reflection of the graph at the diagonal, and the tangent behaves accordingly.
We consider the difference quotient
f
−
1
(
y
)
−
f
−
1
(
b
)
y
−
b
=
f
−
1
(
y
)
−
a
y
−
b
,
{\displaystyle {}{\frac {f^{-1}(y)-f^{-1}(b)}{y-b}}={\frac {f^{-1}(y)-a}{y-b}}\,,}
and have to show that the limit for
y
→
b
{\displaystyle {}y\rightarrow b}
(
y
n
)
n
∈
N
{\displaystyle {}{\left(y_{n}\right)}_{n\in \mathbb {N} }}
sequence
in
E
∖
{
b
}
{\displaystyle {}E\setminus \{b\}}
converging
to
b
{\displaystyle {}b}
fact ,
the function
f
−
1
{\displaystyle {}f^{-1}}
x
n
:=
f
−
1
(
y
n
)
{\displaystyle {}x_{n}:=f^{-1}(y_{n})}
a
{\displaystyle {}a}
x
n
≠
a
{\displaystyle {}x_{n}\neq a}
n
{\displaystyle {}n}
lim
n
→
∞
f
−
1
(
y
n
)
−
a
y
n
−
b
=
lim
n
→
∞
x
n
−
a
f
(
x
n
)
−
f
(
a
)
=
(
lim
n
→
∞
f
(
x
n
)
−
f
(
a
)
x
n
−
a
)
−
1
,
{\displaystyle {}\lim _{n\rightarrow \infty }{\frac {f^{-1}(y_{n})-a}{y_{n}-b}}=\lim _{n\rightarrow \infty }{\frac {x_{n}-a}{f(x_{n})-f(a)}}={\left(\lim _{n\rightarrow \infty }{\frac {f(x_{n})-f(a)}{x_{n}-a}}\right)}^{-1}\,,}
where the right-hand side exists, due to the condition, and the second equation follows from
fact (5) .
◻
{\displaystyle \Box }
