Derivace součtu a rozdílu funkcí [ editovat ] Nechť
f
,
g
∈
C
1
(
R
)
{\displaystyle f,g\in C^{1}(\mathbb {R} )}
(
f
(
x
)
±
g
(
x
)
)
′
=
f
′
(
x
)
±
g
′
(
x
)
.
{\displaystyle (f(x)\pm g(x))'=f'(x)\pm g'(x).}
(
f
(
x
)
±
g
(
x
)
)
′
=
lim
h
→
0
f
(
x
+
h
)
±
g
(
x
+
h
)
−
f
(
x
)
∓
g
(
x
)
h
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
±
lim
h
→
0
g
(
x
+
h
)
−
g
(
x
)
h
=
f
′
(
x
)
±
g
′
(
x
)
{\displaystyle (f(x)\pm g(x))'=\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\pm g(x+h)-f(x)\mp g(x)}{h}}=\lim \limits _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}\pm \lim \limits _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}=f'(x)\pm g'(x)}
Derivace součinu konstanty a funkce [ editovat ] Nechť
f
∈
C
1
(
R
)
{\displaystyle f\in C^{1}(\mathbb {R} )}
c
∈
R
{\displaystyle c\in \mathbb {R} }
(
c
⋅
f
(
x
)
)
′
=
c
⋅
f
′
(
x
)
.
{\displaystyle (c\cdot f(x))'=c\cdot f'(x).}
(
c
⋅
f
(
x
)
)
′
=
lim
h
→
0
c
⋅
f
(
x
+
h
)
−
c
⋅
f
(
x
)
h
=
c
⋅
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
c
⋅
f
′
(
x
)
{\displaystyle (c\cdot f(x))'=\lim \limits _{h\rightarrow 0}{\frac {c\cdot f(x+h)-c\cdot f(x)}{h}}=c\cdot \lim \limits _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}=c\cdot f'(x)}
Derivace součinu funkcí [ editovat ] Nechť
f
∈
C
1
(
R
)
{\displaystyle f\in C^{1}(\mathbb {R} )}
(
f
(
x
)
⋅
g
(
x
)
)
′
=
f
′
(
x
)
⋅
g
(
x
)
+
f
(
x
)
⋅
g
′
(
x
)
.
{\displaystyle (f(x)\cdot g(x))'=f'(x)\cdot g(x)+f(x)\cdot g'(x).}
(
f
(
x
)
⋅
g
(
x
)
)
′
=
lim
h
→
0
f
(
x
+
h
)
⋅
g
(
x
+
h
)
−
f
(
x
)
⋅
g
(
x
)
h
=
lim
h
→
0
f
(
x
+
h
)
⋅
g
(
x
+
h
)
−
f
(
x
)
⋅
g
(
x
+
h
)
+
f
(
x
)
⋅
g
(
x
+
h
)
−
f
(
x
)
⋅
g
(
x
)
h
=
=
lim
h
→
0
f
(
x
+
h
)
⋅
g
(
x
+
h
)
−
f
(
x
)
⋅
g
(
x
+
h
)
h
+
lim
h
→
0
f
(
x
)
⋅
g
(
x
+
h
)
−
f
(
x
)
⋅
g
(
x
)
h
=
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
⋅
lim
h
→
0
g
(
x
+
h
)
+
f
(
x
)
⋅
lim
h
→
0
g
(
x
+
h
)
−
g
(
x
)
h
=
f
′
(
x
)
⋅
g
(
x
)
+
f
(
x
)
⋅
g
′
(
x
)
{\displaystyle {\begin{array}{rcl}(f(x)\cdot g(x))'&=&\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}{h}}=\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x+h)-f(x)\cdot g(x+h)+f(x)\cdot g(x+h)-f(x)\cdot g(x)}{h}}=\\&=&\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x+h)-f(x)\cdot g(x+h)}{h}}+\lim \limits _{h\rightarrow 0}{\frac {f(x)\cdot g(x+h)-f(x)\cdot g(x)}{h}}=\\&=&\lim \limits _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}\cdot \lim \limits _{h\rightarrow 0}g(x+h)+f(x)\cdot \lim \limits _{h\rightarrow 0}{\frac {g(x+h)-g(x)}{h}}=f'(x)\cdot g(x)+f(x)\cdot g'(x)\end{array}}}
Derivace podílu funkcí [ editovat ] Nechť
f
,
g
∈
C
1
(
R
)
{\displaystyle f,g\in C^{1}(\mathbb {R} )}
(
f
(
x
)
g
(
x
)
)
′
=
f
′
(
x
)
⋅
g
(
x
)
−
f
(
x
)
⋅
g
′
(
x
)
g
2
(
x
)
.
{\displaystyle \left({\frac {f(x)}{g(x)}}\right)'={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^{2}(x)}}.}
(
f
(
x
)
g
(
x
)
)
′
=
lim
h
→
0
f
(
x
+
h
)
g
(
x
+
h
)
−
f
(
x
)
g
(
x
)
h
=
lim
h
→
0
f
(
x
+
h
)
⋅
g
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x
)
−
f
(
x
)
⋅
g
(
x
+
h
)
h
⋅
g
(
x
)
⋅
g
(
x
+
h
)
=
=
lim
h
→
0
1
g
(
x
)
⋅
g
(
x
+
h
)
⋅
lim
h
→
0
f
(
x
+
h
)
⋅
g
(
x
+
h
)
−
f
(
x
+
h
)
⋅
g
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x
+
h
)
+
f
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x
+
h
)
⋅
g
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x
)
−
f
(
x
)
⋅
g
(
x
+
h
)
h
=
=
1
g
2
(
x
)
⋅
(
lim
h
→
0
f
(
x
+
h
)
⋅
g
(
x
+
h
)
−
f
(
x
)
⋅
g
(
x
+
h
)
h
−
lim
h
→
0
f
(
x
+
h
)
⋅
g
(
x
+
h
)
−
f
(
x
+
h
)
⋅
g
(
x
)
h
)
=
f
′
(
x
)
⋅
g
(
x
)
−
f
(
x
)
⋅
g
′
(
x
)
g
2
(
x
)
{\displaystyle {\begin{array}{rcl}\left({\frac {f(x)}{g(x)}}\right)'&=&\lim \limits _{h\rightarrow 0}{\frac {{\frac {f(x+h)}{g(x+h)}}-{\frac {f(x)}{g(x)}}}{h}}=\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x)-f(x)\cdot g(x+h)}{h\cdot g(x)\cdot g(x+h)}}=\\&=&\lim \limits _{h\rightarrow 0}{\frac {1}{g(x)\cdot g(x+h)}}\cdot \lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x+h)-f(x+h)\cdot g(x+h)+f(x+h)\cdot g(x)-f(x)\cdot g(x+h)}{h}}=\\&=&{\frac {1}{g^{2}(x)}}\cdot \left(\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x+h)-f(x)\cdot g(x+h)}{h}}-\lim \limits _{h\rightarrow 0}{\frac {f(x+h)\cdot g(x+h)-f(x+h)\cdot g(x)}{h}}\right)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^{2}(x)}}\end{array}}}
Derivace složené funkce [ editovat ] Nechť
f
,
g
∈
C
1
(
R
)
{\displaystyle f,g\in C^{1}(\mathbb {R} )}
(
f
(
g
(
x
)
)
)
′
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
.
{\displaystyle (f(g(x)))'=f'(g(x))\cdot g'(x).}
Položme
y
:=
g
(
x
)
{\displaystyle y:=g(x)}
k
:=
g
(
x
+
h
)
−
g
(
x
)
{\displaystyle k:=g(x+h)-g(x)}
(
f
(
g
(
x
)
)
)
′
=
lim
h
→
0
f
(
g
(
x
+
h
)
)
−
f
(
g
(
x
)
)
h
=
lim
h
→
0
f
(
g
(
x
+
h
)
)
−
f
(
g
(
x
)
)
g
(
x
+
h
)
−
g
(
x
)
⋅
g
(
x
+
h
)
−
g
(
x
)
h
=
=
lim
k
→
0
f
(
y
+
k
)
−
f
(
y
)
k
⋅
g
′
(
x
)
=
f
′
(
y
)
⋅
g
′
(
x
)
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
{\displaystyle {\begin{array}{rcl}(f(g(x)))'&=&\lim \limits _{h\rightarrow 0}{\frac {f(g(x+h))-f(g(x))}{h}}=\lim \limits _{h\rightarrow 0}{\frac {f(g(x+h))-f(g(x))}{g(x+h)-g(x)}}\cdot {\frac {g(x+h)-g(x)}{h}}=\\&=&\lim \limits _{k\rightarrow 0}{\frac {f(y+k)-f(y)}{k}}\cdot g'(x)=f'(y)\cdot g'(x)=f'(g(x))\cdot g'(x)\end{array}}}
Derivace inverzní funkce [ editovat ] Nechť
f
,
g
∈
C
1
(
R
)
{\displaystyle f,g\in C^{1}(\mathbb {R} )}
f
(
g
(
x
)
)
=
id
D
(
g
)
{\displaystyle f(g(x))={\mbox{id}}_{D(g)}}
g
(
f
(
x
)
)
=
id
D
(
f
)
{\displaystyle g(f(x))={\mbox{id}}_{D(f)}}
f
(
x
)
=
y
{\displaystyle f(x)=y}
g
(
y
)
=
x
{\displaystyle g(y)=x}
f
′
(
x
)
=
1
g
′
(
y
)
.
{\displaystyle f'(x)={\frac {1}{g'(y)}}.}
f
′
(
x
0
)
=
lim
x
→
x
0
f
(
x
)
−
f
(
x
0
)
x
−
x
0
=
lim
y
→
y
0
y
−
y
0
g
(
y
)
−
g
(
y
0
)
=
1
lim
y
→
y
0
g
(
y
)
−
g
(
y
0
)
y
−
y
0
=
1
g
′
(
y
0
)
{\displaystyle f'(x_{0})=\lim \limits _{x\rightarrow x_{0}}{\frac {f(x)-f(x_{0})}{x-x_{0}}}=\lim \limits _{y\rightarrow y_{0}}{\frac {y-y_{0}}{g(y)-g(y_{0})}}={\frac {1}{\lim \limits _{y\rightarrow y_{0}}{\frac {g(y)-g(y_{0})}{y-y_{0}}}}}={\frac {1}{g'(y_{0})}}}
Vzorce pro derivování elementárních funkcí [ editovat ] Konstantní funkce [ editovat ] Je-li
f
(
x
)
=
c
{\displaystyle f(x)=c}
c
∈
R
{\displaystyle c\in \mathbb {R} }
f
′
(
x
)
=
0
{\displaystyle f'(x)=0}
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
=
lim
h
→
0
c
−
c
h
=
lim
h
→
0
0
h
=
0
{\displaystyle f'(x)=\lim \limits _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}=\lim \limits _{h\rightarrow 0}{\frac {c-c}{h}}=\lim \limits _{h\rightarrow 0}{\frac {0}{h}}=0}
Mocninná funkce [ editovat ] Je-li
f
(
x
)
=
x
n
{\displaystyle f(x)=x^{n}}
n
∈
N
{\displaystyle n\in \mathbb {N} }
f
′
(
x
)
=
n
⋅
x
n
−
1
{\displaystyle f'(x)=n\cdot x^{n-1}}
f
′
(
x
)
=
lim
h
→
0
(
x
+
h
)
n
−
x
n
h
=
lim
h
→
0
x
n
+
(
n
1
)
x
n
−
1
h
+
…
+
(
n
k
)
x
n
−
k
h
k
+
…
+
h
n
−
x
n
h
=
=
lim
h
→
0
(
n
1
)
x
n
−
1
h
+
…
+
(
n
k
)
x
n
−
k
h
k
+
…
+
h
n
h
=
n
⋅
x
n
−
1
{\displaystyle {\begin{array}{rcl}f'(x)&=&\lim \limits _{h\rightarrow 0}{\frac {(x+h)^{n}-x^{n}}{h}}=\lim \limits _{h\rightarrow 0}{\frac {x^{n}+{\begin{pmatrix}n\\1\end{pmatrix}}x^{n-1}h+\ldots +{\begin{pmatrix}n\\k\end{pmatrix}}x^{n-k}h^{k}+\ldots +h^{n}-x^{n}}{h}}=\\&=&\lim \limits _{h\rightarrow 0}{\frac {{\begin{pmatrix}n\\1\end{pmatrix}}x^{n-1}h+\ldots +{\begin{pmatrix}n\\k\end{pmatrix}}x^{n-k}h^{k}+\ldots +h^{n}}{h}}=n\cdot x^{n-1}\end{array}}}
Je-li
f
(
x
)
=
sin
x
{\displaystyle f(x)=\sin x}
f
′
(
x
)
=
cos
x
{\displaystyle f'(x)=\cos x}
f
′
(
x
)
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
x
h
=
lim
h
→
0
cos
x
⋅
sin
h
+
sin
x
⋅
cos
h
−
sin
x
h
=
lim
h
→
0
cos
x
⋅
sin
h
h
=
cos
x
{\displaystyle f'(x)=\lim \limits _{h\rightarrow 0}{\frac {\sin(x+h)-\sin x}{h}}=\lim \limits _{h\rightarrow 0}{\frac {\cos x\cdot \sin h+\sin x\cdot \cos h-\sin x}{h}}=\lim \limits _{h\rightarrow 0}{\frac {\cos x\cdot \sin h}{h}}=\cos x}
Je-li
f
(
x
)
=
cos
x
{\displaystyle f(x)=\cos x}
f
′
(
x
)
=
−
sin
x
{\displaystyle f'(x)=-\sin x}
f
′
(
x
)
=
lim
h
→
0
cos
(
x
+
h
)
−
cos
x
h
=
lim
h
→
0
cos
x
⋅
cos
h
−
sin
x
⋅
sin
h
−
cos
x
h
=
lim
h
→
0
−
sin
x
⋅
sin
h
h
=
−
sin
x
{\displaystyle f'(x)=\lim \limits _{h\rightarrow 0}{\frac {\cos(x+h)-\cos x}{h}}=\lim \limits _{h\rightarrow 0}{\frac {\cos x\cdot \cos h-\sin x\cdot \sin h-\cos x}{h}}=\lim \limits _{h\rightarrow 0}{\frac {-\sin x\cdot \sin h}{h}}=-\sin x}
Funkce tangens a kotangens [ editovat ] Je-li
f
(
x
)
=
t
g
x
{\displaystyle f(x)=\mathop {\rm {tg\ }} x}
g
(
x
)
=
c
o
t
g
x
{\displaystyle g(x)=\mathop {\rm {cotg\ }} x}
f
′
(
x
)
=
1
cos
2
x
{\displaystyle f'(x)={\frac {1}{\cos ^{2}x}}}
g
′
(
x
)
=
−
1
sin
2
x
{\displaystyle g'(x)=-{\frac {1}{\sin ^{2}x}}}
Přímou aplikací pravidla pro derivaci podílu ihned plyne tvrzení. Například
(
t
g
x
)
′
=
(
sin
x
cos
x
)
′
=
cos
2
x
+
sin
2
x
cos
2
x
=
1
cos
2
x
{\displaystyle (\mathop {\rm {tg\ }} x)'=\left({\frac {\sin x}{\cos x}}\right)'={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}={\frac {1}{\cos ^{2}x}}}
Funkce arkus sinus a arkus kosinus [ editovat ] Je-li
f
(
x
)
=
arcsin
x
{\displaystyle f(x)=\arcsin x}
g
(
x
)
=
arccos
x
{\displaystyle g(x)=\arccos x}
f
′
(
x
)
=
1
1
−
x
2
{\displaystyle f'(x)={\frac {1}{\sqrt {1-x^{2}}}}}
g
′
(
x
)
=
−
1
1
−
x
2
{\displaystyle g'(x)=-{\frac {1}{\sqrt {1-x^{2}}}}}
Označíme-li si
arcsin
x
=
y
{\displaystyle \arcsin x=y}
x
=
sin
y
{\displaystyle x=\sin y}
(
arcsin
x
)
′
=
1
(
sin
y
)
′
=
1
cos
y
=
1
1
−
sin
2
y
=
1
1
−
x
2
{\displaystyle (\arcsin x)'={\frac {1}{(\sin y)'}}={\frac {1}{\cos y}}={\frac {1}{\sqrt {1-\sin ^{2}y}}}={\frac {1}{\sqrt {1-x^{2}}}}}
Analogicky pro
arccos
x
{\displaystyle \arccos x}
Funkce arkus tangens a arkus kotangens [ editovat ] Je-li
f
(
x
)
=
a
r
c
t
g
x
{\displaystyle f(x)=\mathop {\rm {arctg}} x}
g
(
x
)
=
a
r
c
c
o
t
g
x
{\displaystyle g(x)=\mathop {\rm {arccotg}} x}
f
′
(
x
)
=
1
1
+
x
2
{\displaystyle f'(x)={\frac {1}{1+x^{2}}}}
g
′
(
x
)
=
−
1
1
+
x
2
{\displaystyle g'(x)=-{\frac {1}{1+x^{2}}}}
Zcela analogicky předchozím.
Exponenciální funkce [ editovat ] Je-li
f
(
x
)
=
e
x
{\displaystyle f(x)=e^{x}}
f
′
(
x
)
=
e
x
{\displaystyle f'(x)=e^{x}}
f
′
(
x
)
=
lim
h
→
0
e
x
+
h
−
e
x
h
=
e
x
⋅
lim
h
→
0
e
h
−
1
h
{\displaystyle f'(x)=\lim \limits _{h\rightarrow 0}{\frac {e^{x+h}-e^{x}}{h}}=e^{x}\cdot \lim \limits _{h\rightarrow 0}{\frac {e^{h}-1}{h}}}
(
1
+
1
n
+
1
)
n
+
1
≤
e
≤
(
1
+
1
n
−
1
)
n
{\displaystyle \left(1+{\frac {1}{n+1}}\right)^{n+1}\leq e\leq \left(1+{\frac {1}{n-1}}\right)^{n}}
h
∈
R
+
{\displaystyle h\in \mathbb {R} ^{+}}
(
1
+
1
n
+
1
)
(
n
+
1
)
h
≤
e
h
≤
(
1
+
1
n
−
1
)
n
h
{\displaystyle \left(1+{\frac {1}{n+1}}\right)^{(n+1)h}\leq e^{h}\leq \left(1+{\frac {1}{n-1}}\right)^{nh}}
0
<
h
<
1
{\displaystyle 0<h<1}
n
∈
N
{\displaystyle n\in \mathbb {N} }
n
≤
1
h
<
n
+
1
{\displaystyle n\leq {\frac {1}{h}}<n+1}
1
n
+
1
<
h
≤
1
n
{\displaystyle {\frac {1}{n+1}}<h\leq {\frac {1}{n}}}
1
+
1
n
+
1
≤
(
1
+
1
n
+
1
)
(
n
+
1
)
h
≤
e
h
≤
(
1
+
1
n
−
1
)
n
h
≤
1
+
1
n
−
1
{\displaystyle 1+{\frac {1}{n+1}}\leq \left(1+{\frac {1}{n+1}}\right)^{(n+1)h}\leq e^{h}\leq \left(1+{\frac {1}{n-1}}\right)^{nh}\leq 1+{\frac {1}{n-1}}}
1
n
+
1
≤
e
h
−
1
≤
1
n
−
1
{\displaystyle {\frac {1}{n+1}}\leq e^{h}-1\leq {\frac {1}{n-1}}}
1
h
{\displaystyle {\frac {1}{h}}}
1
h
n
+
1
≤
e
h
−
1
h
≤
1
h
n
−
1
{\displaystyle {\frac {\frac {1}{h}}{n+1}}\leq {\frac {e^{h}-1}{h}}\leq {\frac {\frac {1}{h}}{n-1}}}
1
−
1
n
+
1
=
n
n
+
1
≤
1
h
n
+
1
≤
e
h
−
1
h
≤
1
h
n
−
1
≤
n
+
1
n
−
1
=
1
+
2
n
−
1
{\displaystyle 1-{\frac {1}{n+1}}={\frac {n}{n+1}}\leq {\frac {\frac {1}{h}}{n+1}}\leq {\frac {e^{h}-1}{h}}\leq {\frac {\frac {1}{h}}{n-1}}\leq {\frac {n+1}{n-1}}=1+{\frac {2}{n-1}}}
n
→
∞
{\displaystyle n\rightarrow \infty }
h
→
0
+
{\displaystyle h\rightarrow 0_{+}}
lim
h
→
0
+
e
h
−
1
h
=
1
{\displaystyle \lim \limits _{h\rightarrow 0_{+}}{\frac {e^{h}-1}{h}}=1}
h
∈
R
−
{\displaystyle h\in \mathbb {R} ^{-}}
Funkce přirozený logaritmus [ editovat ] Je-li
f
(
x
)
=
ln
x
{\displaystyle f(x)=\ln x}
f
′
(
x
)
=
1
x
{\displaystyle f'(x)={\frac {1}{x}}}
(ln e(x))' = 1/(x*ln (e))
Zcela analogicky předchozím důkazům pro derivaci inverzních funkcí.
