\(\displaystyle{ \Large f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}}\)
I jeszcze kilka przydatnych wzorów:
- Pochodna iloczynu funkcji i liczby:
\(\displaystyle{ \left( a\cdot f \left( x \right) \right) '=\lim\limits_{h\to0}\frac{a\cdot f \left( x+h \right) -a\cdot f \left( x \right) }{h}=a\cdot\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=a\cdot f' \left( x \right)}\) - Pochodna sumy:
\(\displaystyle{ \left( f \left( x \right) +g \left( x \right) \right) '=\lim\limits_{h\to0}\frac{f \left( x+h \right) +g \left( x+h \right) - \left( f \left( x \right) +g \left( x \right) \right) }{h}= \lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}+\lim\limits_{h\to0}\frac{g \left( x+h \right) -g \left( x \right) }{h}=f' \left( x \right) + g' \left( x \right)}\) - Pochodna różnicy:
\(\displaystyle{ \left( f \left( x \right) - g \left( x \right) \right) '=\lim\limits_{h\to0}\frac{f \left( x+h \right) -g \left( x+h \right) - \left( f \left( x \right) -g \left( x \right) \right) }{h}= \lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}-\lim\limits_{h\to0}\frac{g \left( x+h \right) -g \left( x \right) }{h}=f' \left( x \right) - g' \left( x \right)}\) - Pochodna iloczynu:
\(\displaystyle{ \left( f \left( x \right) \cdot g \left( x \right) \right) '=\lim\limits_{h\to0}\frac{f \left( x+h \right) g \left( x+h \right) -f \left( x \right) g \left( x \right) }{h}=\\=\lim\limits_{h\to0}\frac{f \left( x+h \right) g \left( x+h \right) -f \left( x \right) g \left( x \right) +f \left( x \right) g \left( x+h \right) -f \left( x \right) g \left( x+h \right) }{h}=\\=\lim\limits_{h\to0}\frac{f \left( x \right) \left( g \left( x+h \right) -g \left( x \right) \right) +g \left( x+h \right) \left( f \left( x+h \right) -f \left( x \right)\right) }{h}= f \left( x \right) \cdot\lim\limits_{h\to0}\frac{g \left( x+h \right) -g \left( x \right) }{h}+g \left( x \right) \cdot\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}= f \left( x \right) \cdot g' \left( x \right) +f' \left( x \right) \cdot g \left( x \right) }\) - Pochodna ilorazu:
\(\displaystyle{ \left( \frac{f \left( x \right) }{g \left( x \right) } \right) '=\lim\limits_{h\to0}\frac{\frac{f \left( x+h \right) }{g \left( x+h \right) }-\frac{f \left( x \right) }{g \left( x \right) }}{h}= \lim\limits_{h\to0}\frac{f \left( x+h \right) g \left( x \right) -f \left( x \right) g \left( x+h \right) }{h\cdot g \left( x \right) g \left( x+h \right) }= \\=\lim\limits_{h\to0}\frac{f \left( x+h \right) g \left( x \right) -f \left( x \right) g \left( x+h \right) +f \left( x \right) g \left( x \right) -f \left( x \right) g \left( x \right) }{h\cdot g \left( x \right) g \left( x \right) }= \\=\lim\limits_{h\to0}\frac{g \left( x \right) \left( f \left( x+h \right) -f \left( x \right) \right) -f \left( x \right) \left( g \left( x+h \right) -g \left( x \right) \right) }{h\cdot \left( g \left( x \right) \right) ^2}= \frac{g \left( x \right) }{ \left( g \left( x \right) \right) ^2}\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}-\frac{f \left( x \right) }{ \left( g \left( x \right) \right) ^2}\lim\limits_{h\to0}\frac{g \left( x+h \right) -g \left( x \right) }{h}= \frac{f' \left( x \right) g \left( x \right) -f \left( x \right) g' \left( x \right) }{ \left( g \left( x \right) ^2 \right) },\:\; gdy\,g \left( x \right) \neq0}\)
POCHODNE WAŻNIEJSZYCH FUNKCJI:
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{a-a}{h}=0}\)
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{a \left( x+h \right) +b- \left( ax+b \right) }{h}=\lim\limits_{h\to0}\frac{ax+ah+b-ax-b}{h}= \lim\limits_{h\to0}\frac{ah}{h}=a}\)
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}= \lim\limits_{h\to0}\frac{x+h-x}{h \left( \sqrt{x+h}+\sqrt{x} \right) }=\lim\limits_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}}\)
- \(\displaystyle{ f' \left( x \right) =f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=f' \left( x \right) =\lim\limits_{h\to0}\frac{\frac{a}{x+h}-\frac{a}{x}}{h}= f' \left( x \right) =\lim\limits_{h\to0}\frac{ax-a \left( x+h \right) }{h\cdot x \left( x+h \right) }= \lim\limits_{h\to0}\frac{-a}{x \left( x+h \right) }=-\frac{a}{x^2}}\)
-
\(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}= \lim\limits_{h\to0}\frac{a^{x+h}-a^x}{h}=\lim\limits_{h\to0}\frac{a^x \left( a^h-1 \right) }{h}=a^x\lim\limits_{h\to0}\frac{a^h-1}{h}}\)
Zastosujemy teraz podstawienie \(\displaystyle{ a^h-1=z}\). Jeśli \(\displaystyle{ h\to0}\), to \(\displaystyle{ a^h\to1\,\Rightarrow\,z\to0}\).
\(\displaystyle{ a^h=z+1\,\Leftrightarrow\,h=\log _a \left( z+1 \right)}\)
\(\displaystyle{ f' \left( x \right) =a^x\lim\limits_{h\to0}\frac{a^h-1}{h}=a^x\lim\limits_{z\to0}\frac{z}{\log _a \left( z+1 \right) }= a^x\lim\limits_{z\to0}\frac{1}{\frac{1}{z}\log _a \left( z+1 \right) }=a^x\lim\limits_{z\to0}\frac{1}{\log _a \left( 1+z \right) ^{\frac{1}{z}}}= a^x\cdot\frac{1}{\log _a \left( \lim\limits_{z\to0} \left( 1+z \right) ^{\frac{1}{z}} \right) }=a^x\cdot\frac{1}{\log _ae}=a^x\ln {a}}\) -
szczególny przypadek powyższego wzoru
\(\displaystyle{ f' \left( x \right) =e^x\ln {e}=e^x}\) - \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{\log _a \left( x+h \right) -\log _ax}{h}= \lim\limits_{h\to0}\frac{1}{h}\log _a \left( \frac{x+h}{x} \right) = {\lim\limits_{h\to0}\log _a \left( 1+\frac{h}{x} \right) ^{\frac{1}{h}}}=\log _a \left( \lim\limits_{h\to0} \left( 1+\frac{h}{x} \right) ^{\frac{1}{h}} \right) = \log _a \left( \lim\limits_{h\to0} \left( \left( 1+\frac{h}{x} \right) ^{\frac{x}{h}} \right) ^{\frac{1}{x}} \right) =\log _ae^{\frac{1}{x}}=\frac{1}{x}\log _ae=\frac{1}{x\ln {a}}}\)
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szczególny przypadek powyższego wzoru
\(\displaystyle{ f' \left( x \right) =\frac{1}{x\ln {e}}=\frac{1}{x}}\) - \(\displaystyle{ f' \left( x \right) = \left( x^a \right) '= \left( e^{\ln {x^a}} \right) '= \left( e^{a\ln {x}} \right) '=e^{a\ln {x}}\cdot \left( a\ln {x} \right) '= e^{\ln {x^a}}\cdot a\frac{1}{x}=x^a\cdot a\frac{1}{x}=a\frac{x^a}{x}=a\cdot x^{a-1}}\)
- \(\displaystyle{ f' \left( x \right) = \left( x^x \right) '= \left( e^{\ln {x^x}} \right) '= \left( e^{x\ln {x}} \right) '=e^{x\ln {x}}\cdot \left( x\ln {x} \right) '= e^{\ln {x^x}}\cdot \left( 1\cdot\ln {x}+x\cdot\frac{1}{x} \right) =x^x \left( \ln {x}+1 \right)}\)
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{\sin { \left( x+h \right) }-\sin {x}}{h}= \lim\limits_{h\to0}\frac{2\sin { \left( \frac{x+h-x}{2} \right) }\cos { \left( \frac{x+h+x}{2} \right) }}{h}= \lim\limits_{h\to0}\frac{\sin { \left( \frac{h}{2} \right) }\cos { \left( x+\frac{h}{2} \right) }}{\frac{h}{2}}= \cos {x}\cdot\lim\limits_{h\to0}\frac{\sin { \left( \frac{h}{2} \right) }}{\frac{h}{2}}=\cos {x}}\)
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{\cos { \left( x+h \right) }-\cos {x}}{h}= \lim\limits_{h\to0}\frac{-2\sin { \left( \frac{x+h+x}{2} \right) }\sin { \left( \frac{x+h-x}{2} \right) }}{h}= \lim\limits_{h\to0}\frac{-\sin { \left( \frac{h}{2} \right) }\sin { \left( x+\frac{h}{2} \right) }}{\frac{h}{2}}= -\sin {x}\cdot\lim\limits_{h\to0}\frac{\sin { \left( \frac{h}{2} \right) }}{\frac{h}{2}}=-\sin {x}}\)
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{\tan { \left( x+h \right) }-\tan {x}}{h}= \lim\limits_{h\to0}\frac{\frac{\sin { \left( x+h-x \right) }}{\cos { \left( x+h \right) }\cos {x}}}{h}= \lim\limits_{h\to0}\frac{\sin {h}}{h\cos { \left( x+h \right) }\cos {x}}=\frac{1}{\cos ^2{x}}\cdot\lim\limits_{h\to0}\frac{\sin {h}}{h}=\frac{1}{\cos ^2{x}}}\)
- \(\displaystyle{ f' \left( x \right) =\lim\limits_{h\to0}\frac{f \left( x+h \right) -f \left( x \right) }{h}=\lim\limits_{h\to0}\frac{\cot { \left( x+h \right) }-\cot {x}}{h}= \lim\limits_{h\to0}\frac{\frac{\sin { \left( x- \left( x+h \right) \right) }}{\sin { \left( x+h \right) }\sin {x}}}{h}= \lim\limits_{h\to0}\frac{-\sin {h}}{h\sin { \left( x+h \right) }\sin {x}}=-\frac{1}{\sin ^2{x}}\cdot\lim\limits_{h\to0}\frac{\sin {h}}{h}=-\frac{1}{\sin ^2{x}}}\)
18 lis 2012, miki999 - skalowanie nawiasów