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Теория: 01 Нахождение производной (таблица производных)

Задание

\(\displaystyle 1)\,(c)^{\prime}=0, \, c=const\)

\(\displaystyle 2)\,(x^n)^{\prime}=nx^{n-1}\)

\(\displaystyle 3)\,(a^x)^{\prime}=a^x \cdot \ln{a}\)

\(\displaystyle 4)\,(e^x)^{\prime}=e^x\)

\(\displaystyle 5)\,(\log_a{x})^{\prime}=\frac{1}{x\ln{a}}\)

\(\displaystyle 6)\,(\ln{x})^{\prime}=\frac{1}{x}\)

\(\displaystyle 7)\,(\sin{x})^{\prime}=\cos{x}\)

\(\displaystyle 8)\,(\cos{x})^{\prime}=-\sin{x}\)

\(\displaystyle 9)\,(\tg{x})^{\prime}=\frac{1}{{\cos^2{x}}}\)

\(\displaystyle 10)\,(\ctg{x})^{\prime}=-\frac{1}{{\sin^2{x}}}\)

\(\displaystyle 11)\,(\sqrt{x})^{\prime}=\frac{1}{2\sqrt{x}}\)

\(\displaystyle 12)\,(\arcsin{x})^{\prime}=\frac{1}{\sqrt{1-x^2}}\)

\(\displaystyle 13)\,(\arccos{x})^{\prime}=-\frac{1}{\sqrt{1-x^2}}\)

\(\displaystyle 14)\,(\arctg{x})^{\prime}=\frac{1}{1+x^2}\)

\(\displaystyle 15)\,(\arcctg{x})^{\prime}=-\frac{1}{1+x^2}\)

Решение