Sigui c un nombre real .
Es considera la funció constant f de valor c :
∀
x
∈
R
,
∀
h
∈
R
∗
,
f
(
x
+
h
)
−
f
(
x
)
h
=
c
−
c
h
=
0
{\displaystyle \forall x\in \mathbb {R} ,\forall h\in \mathbb {R^{*}} ,{\frac {f(x+h)-f(x)}{h}}={\frac {c-c}{h}}=0}
per tant
∀
x
∈
R
,
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
0
{\displaystyle \forall x\in \mathbb {R} ,f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}=0}
Així la derivada d'una funció constant és la funció nul·la.
Sigui la funció f :
f
(
x
)
=
x
n
{\displaystyle f(x)=x^{n}\,}
I
{\displaystyle {I}\,}
h
≠
0
{\displaystyle h\not =0}
∀
a
∈
I
,
∀
(
a
+
h
)
∈
I
{\displaystyle \forall a\in {I},\forall {(a+h)}\in {I}\,}
t
(
h
)
=
f
(
a
+
h
)
−
f
(
a
)
h
{\displaystyle t(h)={\frac {f(a+h)-f(a)}{h}}}
t
(
h
)
=
(
a
+
h
)
n
−
a
n
h
{\displaystyle t(h)={\frac {(a+h)^{n}-a^{n}}{h}}}
t
(
h
)
=
(
a
n
+
n
a
n
−
1
h
+
p
3
a
n
−
2
h
2
+
p
4
a
n
−
3
h
3
.
.
.
p
n
a
h
n
−
1
+
p
n
+
1
h
n
)
−
a
n
h
{\displaystyle t(h)={\frac {(a^{n}+na^{n-1}h+p_{3}a^{n-2}h^{2}+p_{4}a^{n-3}h^{3}...p_{n}ah^{n-1}+p_{n+1}h^{n})-a^{n}}{h}}}
On els coeficients
p
i
{\displaystyle p_{i}}
triangle de Tartaglia (
p
1
=
1
{\displaystyle p_{1}=1}
p
2
=
n
{\displaystyle p_{2}=n}
a
n
{\displaystyle a^{n}}
h
{\displaystyle h}
t
(
h
)
=
n
a
n
−
1
+
p
3
a
n
−
2
h
+
p
4
a
n
−
3
h
2
.
.
.
p
n
a
h
n
−
2
+
p
n
+
1
h
n
−
1
{\displaystyle t(h)=na^{n-1}+p_{3}a^{n-2}h+p_{4}a^{n-3}h^{2}...p_{n}ah^{n-2}+p_{n+1}h^{n-1}\,}
f
′
(
a
)
=
lim
h
→
0
t
(
h
)
=
n
a
n
−
1
{\displaystyle f'(a)=\lim _{h\rightarrow 0}t(h)=na^{n-1}}
Es considera la funció f definida sobre
R
{\displaystyle \mathbb {R} }
∀
x
∈
R
,
f
(
x
)
=
x
2
{\displaystyle \forall x\in \mathbb {R} ,f(x)=x^{2}}
∀
x
∈
R
,
∀
h
∈
R
∗
,
f
(
x
+
h
)
−
f
(
x
)
h
=
(
x
+
h
)
2
−
x
2
h
{\displaystyle \forall x\in \mathbb {R} ,\forall h\in \mathbb {R} ^{*},{\frac {f(x+h)-f(x)}{h}}={\frac {(x+h)^{2}-x^{2}}{h}}}
=
(
x
+
h
−
x
)
(
x
+
h
+
x
)
h
=
h
(
2
x
+
h
)
h
=
2
x
+
h
{\displaystyle ={\frac {(x+h-x)(x+h+x)}{h}}={\frac {h(2x+h)}{h}}=2x+h}
per tant
f
′
(
x
)
=
lim
h
→
0
(
2
x
+
h
)
=
2
x
{\displaystyle f'(x)=\lim _{h\rightarrow 0}(2x+h)=2x}
la derivada de f és per tant la funció f' definida per
∀
x
∈
R
,
f
′
(
x
)
=
2
x
{\displaystyle \forall x\in \mathbb {R} ,f'(x)=2x}
Es considera la funció f =√x
∀
x
∈
R
+
∗
,
∀
h
∈
R
∗
,
h
>
−
x
,
f
(
x
+
h
)
−
f
(
x
)
h
=
x
+
h
−
x
h
{\displaystyle \forall x\in \mathbb {R} _{+}^{*},\forall h\in \mathbb {R} ^{*},h>-x,\quad {\frac {f(x+h)-f(x)}{h}}={\frac {{\sqrt {x+h}}-{\sqrt {x}}}{h}}}
=
(
x
+
h
−
x
)
(
x
+
h
+
x
)
h
(
x
+
h
+
x
)
{\displaystyle ={\frac {({\sqrt {x+h}}-{\sqrt {x}})({\sqrt {x+h}}+{\sqrt {x}})}{h({\sqrt {x+h}}+{\sqrt {x}})}}}
=
x
+
h
−
x
h
(
x
+
h
+
x
)
=
1
x
+
h
+
x
{\displaystyle ={\frac {x+h-x}{h({\sqrt {x+h}}+{\sqrt {x}})}}={\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}}
per tant
∀
x
∈
R
+
∗
,
f
′
(
x
)
=
lim
h
→
0
1
x
+
h
+
x
=
1
2
x
{\displaystyle \forall x\in \mathbb {R} _{+}^{*},f'(x)=\lim _{h\rightarrow 0}{\frac {1}{{\sqrt {x+h}}+{\sqrt {x}}}}={\frac {1}{2{\sqrt {x}}}}}
D'altra banda,
∀
h
∈
R
+
∗
,
f
(
h
)
−
f
(
0
)
h
=
h
h
=
1
h
{\displaystyle \forall h\in \mathbb {R} _{+}^{*},{\frac {f(h)-f(0)}{h}}={\frac {\sqrt {h}}{h}}={\frac {1}{\sqrt {h}}}}
lim
h
→
0
f
(
h
)
−
f
(
0
)
h
=
+
∞
{\displaystyle \lim _{h\rightarrow 0}{\frac {f(h)-f(0)}{h}}=+\infty }
per tant f no és derivable en 0 i la seva gràfica admet en 0 una semi tangent vertical.
Heus aquí una sèrie d'exemples de derivades calculades a partir de les fórmules establertes pel mètode amb el límit.
Es consideren les funcions següents i tot seguit es presenta el procés de càlcul de les seves derivades:
1.
y
=
x
2
+
5
x
−
3
{\displaystyle y=x^{2}+5x-3\,}
2.
y
=
3
x
2
−
9
x
+
2
3
{\displaystyle y=3x^{2}-9x+{\frac {2}{3}}\,}
3.
y
=
−
4
x
2
+
4
7
x
−
1
{\displaystyle y=-4x^{2}+{\frac {4}{7}}x-1\,}
Derivació:
1.
y
=
x
2
+
5
x
−
3
{\displaystyle y=x^{2}+5x-3\,}
y
′
=
(
x
2
)
′
+
(
5
x
)
′
−
(
3
)
′
{\displaystyle y'=(x^{2})'+(5x)'-(3)'\,}
y
′
=
2
x
+
5
+
0
{\displaystyle y'=2x+5+0\,}
y
′
=
2
x
+
5
{\displaystyle y'=2x+5\,}
2.
y
=
3
x
2
−
9
x
+
2
3
{\displaystyle y=3x^{2}-9x+{\frac {2}{3}}\,}
y
′
=
(
3
x
2
)
′
−
(
9
x
)
′
+
(
2
3
)
′
{\displaystyle y'=(3x^{2})'-(9x)'+({\frac {2}{3}})'\,}
y
′
=
6
x
−
9
+
0
{\displaystyle y'=6x-9+0\,}
y
′
=
6
x
−
9
{\displaystyle y'=6x-9\,}
3.
y
=
−
4
x
2
+
4
7
x
−
1
{\displaystyle y=-4x^{2}+{\frac {4}{7}}x-1\,}
y
′
=
(
−
4
x
2
)
′
+
(
4
7
x
)
′
−
(
1
)
′
{\displaystyle y'=(-4x^{2})'+({\frac {4}{7}}x)'-(1)'\,}
y
′
=
−
8
x
+
4
7
−
0
{\displaystyle y'=-8x+{\frac {4}{7}}-0\,}
y
′
=
−
8
x
+
4
7
{\displaystyle y'=-8x+{\frac {4}{7}}\,}
Es consideren les funcions següents i tot seguit es presenta el procés de càlcul de les seves derivades:
1.
y
=
2
x
3
+
6
x
2
−
4
x
+
9
π
{\displaystyle y=2x^{3}+6x^{2}-4x+{\frac {9}{\pi }}\,}
2.
y
=
−
x
3
−
5
x
2
+
2
3
x
−
1
{\displaystyle y=-x^{3}-5x^{2}+{\frac {2}{3}}x-1\,}
3.
y
=
5
17
x
3
+
x
2
−
2
x
+
e
{\displaystyle y={\frac {5}{17}}x^{3}+x^{2}-2x+e\,}
Derivades:
1.
y
=
2
x
3
+
6
x
2
−
4
x
+
9
π
{\displaystyle y=2x^{3}+6x^{2}-4x+{\frac {9}{\pi }}\,}
y
′
=
(
2
x
3
)
′
+
(
6
x
2
)
′
−
(
4
x
)
′
+
(
9
π
)
′
{\displaystyle y'=(2x^{3})'+(6x^{2})'-(4x)'+\left({\frac {9}{\pi }}\right)'\,}
y
′
=
6
x
2
+
12
x
−
4
+
0
{\displaystyle y'=6x^{2}+12x-4+0\,}
y
′
=
2
(
3
x
2
+
6
x
−
2
)
{\displaystyle y'=2(3x^{2}+6x-2)\,}
2.
y
=
−
x
3
−
5
x
2
+
2
3
x
−
1
{\displaystyle y=-x^{3}-5x^{2}+{\frac {2}{3}}x-1\,}
y
′
=
−
(
x
3
)
′
−
(
5
x
2
)
′
+
(
2
3
x
)
′
−
(
1
)
′
{\displaystyle y'=-(x^{3})'-(5x^{2})'+\left({\frac {2}{3}}x\right)'-(1)'\,}
y
′
=
−
3
x
2
−
10
x
+
2
3
−
0
{\displaystyle y'=-3x^{2}-10x+{\frac {2}{3}}-0\,}
y
′
=
−
3
x
2
−
10
x
+
2
3
{\displaystyle y'=-3x^{2}-10x+{\frac {2}{3}}\,}
3.
y
=
5
17
x
3
+
x
2
−
2
x
+
e
{\displaystyle y={\frac {5}{17}}x^{3}+x^{2}-2x+e\,}
y
′
=
(
5
17
x
3
)
′
+
(
x
2
)
′
−
(
2
x
)
′
+
(
e
)
′
{\displaystyle y'=\left({\frac {5}{17}}x^{3}\right)'+(x^{2})'-(2x)'+(e)'\,}
y
′
=
5
×
3
x
2
17
+
2
x
−
2
+
0
{\displaystyle y'={\frac {5\times 3x^{2}}{17}}+2x-2+0\,}
y
′
=
15
x
2
17
+
2
x
−
2
{\displaystyle y'={\frac {15x^{2}}{17}}+2x-2\,}
Sia la funció y :
y
(
x
)
=
a
x
b
a
≠
0
,
b
∈
R
{\displaystyle y(x)=ax^{b}\qquad a\not =0,b\in \mathbb {R} }
Llavors, la derivada n-èsima de y ve donada, sobre intervals convenients, per :
∀
n
∈
N
∗
:
y
(
n
)
(
x
)
=
a
∏
k
=
0
n
−
1
(
b
−
k
)
⋅
x
b
−
n
{\displaystyle \forall n\in \mathbb {N} ^{*}\qquad :\qquad y^{(n)}(x)=a\prod _{k=0}^{n-1}(b-k)\cdot x^{b-n}}
Viccionari
