Zadania z pochodnych i rozwiązania
Zadanie z całek (i rozwiązania przez podstawienie i przez części)
| POCHODNE | CAŁKI |
| $$ (C)’ = 0 $$ $$ (ax)’ = a $$ | $$\int a dx = ax + C$$ $$\int (ax + b) dx = \frac{1}{2}ax^2 + bx + C$$ |
| $$ (x^n)’ = n \cdot x^{n-1} $$ $$ x’ = 1 $$ | $$\int x^n dx = \frac{1}{n+1}x^{n+1} + C, \quad (n \neq -1)$$ $$\int dx = x + C$$ |
| $$ (e^x)’ = e^x $$ $$ (e^{ax})’ = ae^{ax} $$ $$ (a^x)’ = a^x lna $$ | $$\int e^x dx = e^x + C$$ $$ \int e^{ax} dx = \frac{1}{a}e^{ax} + C $$ $$\int a^x dx = \frac{a^x}{\ln a} + C$$ |
| $$ (\log_a x)’ = \frac{1}{x \ln a} $$ $$ (lnx)’=\frac{1}{x} $$ | $$\int \frac{1}{x} dx = \ln |x| + C$$ |
| $$ (\sin x)’ = \cos x $$ $$ (\sin (ax))’ = a\cos(ax) $$ | $$\int \cos(x) dx = \sin(x) + C$$ $$\int \cos(ax) dx = \frac{1}{a}\sin(ax) + C $$ |
| $$ (\cos x)’ = -\sin x $$ $$ (\cos (ax))’ = -a\sin(ax) $$ | $$\int \sin(x) dx = -\cos(x) + C$$ $$\int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C$$ |
| $$ (\tan x)’ = \frac{1}{cos^2 x }$$ | $$ \int \frac{1}{cos^2(x)} dx = \tan x + C $$ $$\int \tan x dx = -\ln |\cos x| + C )$$ $$ \int \tan(ax) dx = -\frac{1}{a}\ln |\cos(ax)|+C$$ |
| $$ (\cot x)’ = -\frac{1}{sin^2 x} $$ | $$ \int \frac{1}{sin^2(x) } dx = -\cot x + C $$ $$\int \cot x dx = \ln |\sin x| + C $$ $$\int \cot(ax) dx = \frac{1}{a}\ln |\sin(ax)| + C $$ |
| $$(\arcsin x)’ = \frac{1}{\sqrt{1-x^2}} $$ | $$ \int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C $$ |
| $$(\arccos x)’ = -\frac{1}{\sqrt{1-x^2}} $$ | $$ \int \frac{-1}{\sqrt{1-x^2}} dx = \arccos x + C $$ |
| $$ (\arctan x)’ = \frac{1}{1+x^2} $$ | $$\int \frac{1}{1+x^2} dx = \arctan x + C $$ |
| $$(\text{arccot} x)’ = -\frac{1}{1+x^2} $$ | $$\int \frac{-1}{1+x^2} dx = \text{arccot} x + C $$ |
| $$(\sinh x)’ = \cosh x $$ | $$ \int \cosh(x) dx = \sinh(x)+C$$ |
| $$(\cosh x)’ = \sinh x $$ | $$ \int \sinh(x) dx = \cosh(x) + C$$ |
| $$( \text{tanh}x)’= 1 – \tanh^2(x)$$ | $$ \int \frac{1}{\cosh^2(x)} dx = \tanh(x) +C$$ |
| $$f(x) + g(x))’ = f'(x) + g'(x)$$ $$(f(x) – g(x))’ = f'(x) – g'(x)$$ | |
| $$ (f(x) \cdot g(x))’ = f'(x) \cdot g(x) + f(x) \cdot g'(x) $$ | |
$$ \left(\frac{f(x)}{g(x)}\right)’ = \frac{f'(x) \cdot g(x) – f(x) \cdot g'(x)}{g(x)^2} $$ | |
| Pochodna funkcji złożonej (reguła łańcuchowa): $ (f(g(x)))’ = f'(g(x)) \cdot g'(x) $ | |
| Pochodna funkcji odwrotnej: $ (f^{-1}(y))’ = \frac{1}{f'(f^{-1}(y))}$ | |