-----------------------------------------------POCHODNE--------------------------------------------------

sinx^′ = cosx

cosx^′ = −sinx

$$\text{tg}x^{'} = \frac{1}{\text{cosx}^{2}}$$

$$\text{ctg}x^{'} = \frac{- 1}{\text{sinx}^{2}}$$

$$\arcsin x^{'} = \frac{1}{\sqrt{1 - x^{2}}}$$

$$\arccos x^{'} = \frac{- 1}{\sqrt{1 - x^{2}}}$$

$$\text{arctg}x^{'} = \frac{1}{1 + x^{2}}$$

$$\text{arcctg}x^{'} = \frac{- 1}{1 + x^{2}}$$

e^x′ = e^x

a^x = loga * a^x,       a > 0

$$\ln x^{'} = \frac{1}{x}$$

$$\operatorname{}x = \frac{1}{\text{xlna}}$$

x^α′ = αx^α − 1

----------------------------------------Działania na pochodnych:------------------------------------------

$${\ln\left( f\left( x \right) \right)}^{'} = \frac{f^{'}\left( x \right)}{f\left( x \right)}$$

(f(x)^g(x))^′ = (e^lnf(x)g(x))^′ = e^{g(x) * lnf(x)}

$$\left( \frac{f\left( x \right)}{g\left( x \right)} \right)^{'} = \frac{f^{'}\left( x \right)g\left( x \right) - f\left( x \right)g^{'}\left( x \right)}{{g\left( x \right)}^{2}}$$

(f(x)g(x))^′ = f^′(x)g(x) + f(x)g^′(x)

[f(y)]^′ = f(y)^′ * y′

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----------------------------------------------------CAŁKI----------------------------------------------------

$$\int_{}^{}{x^{a}dx = \frac{x^{a + 1}}{a + 1} + C}$$

$$\int_{}^{}{\frac{\text{dx}}{x} = lnx + C}$$

∫e^xdx = e^x + C

$$\int_{}^{}{a^{x}dx = \frac{a^{x}}{\text{lna}}} + C$$

∫sinx dx =   − cosx + C

∫tgxdx = −ln|cox| + C

$$\int_{}^{}{\frac{\text{dx}}{\text{cosx}^{2}} = \text{tgx}^{2} + C}$$

$$\int_{}^{}{\frac{\text{dx}}{\text{sinx}^{2}} = - ctgx + C}$$

--------------------------------------------Działania na całkach:-------------------------------------------

$$\int_{}^{}{\frac{f^{'}\left( x \right)}{f\left( x \right)}dx = \ln\left| f\left( x \right) \right| + C}$$

n=2,3,4,5….

$$\int_{}^{}{\frac{\text{dx}}{\left( x^{2} + a^{2} \right)^{n}} = \frac{1}{2n - 2}*\frac{x}{a^{2}\left( a^{2} + x^{2} \right)^{n - 1}} + \frac{2n - 3}{2n - 2}*\frac{1}{a^{2}}\int_{}^{}\frac{\text{dx}}{\left( a^{2} + x^{2} \right)^{n - 1}}}$$

------------------------------------------Całkowanie przez części:----------------------------------------

∫f(x)g^′(x) = f(x)g(x) − ∫f^′(x)g(x)dx

-----------------------------------------Rozkład na ułamki proste:----------------------------------------

$$\int_{}^{}{\frac{W\left( x \right)}{P\left( x \right)} = \int_{}^{}\frac{a_{k}}{\left( x - x_{0} \right)^{k}} + \int_{}^{}\frac{a_{k - 1}}{\left( x - x_{0} \right)^{k - 1}} + \ldots + \int_{}^{}\frac{a_{1}}{\left( x - x_{0} \right)} + \int_{}^{}\frac{b_{1}}{\left( x - x_{1} \right)}}$$

np.

$$\frac{4}{x^{2}(x - 2)} = \frac{a_{2}}{{(x - 0)}^{2}} + \frac{a_{1}}{x - 0} + \frac{b}{(x - 2)}$$

$$\frac{4}{x^{2}\left( x - 2 \right)} = \frac{a_{2}\left( x - 2 \right) + a_{1}x\left( x - 2 \right) + bx^{2}}{x^{2}\left( x - 2 \right)} = \frac{x^{2}\left( a_{1} + b \right) + x\left( a_{2} - 2a_{1} \right) - 2a_{2}}{x^{2}\left( x - 2 \right)}$$

$$\left\{ \begin{matrix} a_{1} + b = 0 \\ a_{2} - 2a_{1} = 0 \\ - 2a_{2} = 4 \\ \end{matrix} \right.\ $$

$$\frac{4}{x^{2}(x - 2)} = \frac{- 2}{x^{2}} + \frac{- 1}{x} + \frac{1}{x - 2}$$

$$\int_{}^{}\frac{4}{x^{2}(x - 2)} = \int_{}^{}\frac{- 2}{x^{2}} + \int_{}^{}\frac{- 1}{x} + \int_{}^{}\frac{1}{x - 2}$$

--------------------------------------------------Podstawienie:----------------------------------------------

$$\int_{}^{}{\frac{\sqrt[n]{\frac{ax + b}{px + q}}}{{(c - x)}^{2}}\text{dx}} = \left\{ \begin{matrix} \sqrt[n]{\frac{ax + b}{px + q}} = t \\ x = \frac{costam}{costam} \\ dx = \left( \frac{costam}{costam} \right)'dt \\ \end{matrix} \right.\ = \int_{}^{}{\frac{t}{\left( c - \frac{costam}{costam} \right)^{2}}*}(\frac{costam}{costam})'dt$$

---------------------------------------Do podstawienia: (NA BLACHE!)--------------------------------

$$\text{tg}\frac{x}{2} = t$$

$$sinx = \frac{2t}{1 + t^{2}}$$

$$cosx = \frac{1 - t^{2}}{1 + t^{2}}$$

$$dx = \frac{2}{1 + t^{2}}$$

ZASTOSOWANIE CAŁEK:

---------------------------------------------Długość krzywej:-----------------------------------------------

y(x),x∈[a,b]

$$L = \int_{a}^{b}{\sqrt{1 + {(y(x)')}^{2}}\text{dx}}$$

x=x(t), y=y(t), a≤t≤b

$$L = \int_{a}^{b}{\sqrt{{(x')}^{2} + {(y')}^{2}}\text{dt}}$$

$$L = \int_{a}^{b}\sqrt{\left( x^{'} \right)^{2} + \left( y^{'} \right)^{2} + \left( z^{'} \right)^{2}}\text{dt}$$

----------------------------------------------Objętość figury:-----------------------------------------------

V = π∫_a^by(x)²dx

------------------------------------------Pole powierzchni figury:-----------------------------------------

$$S = 2\pi\int_{a}^{b}{y(x)\sqrt{1 + {y(x)}^{2}}}\text{dx}$$

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------------------------------------------------GRANICE:---------------------------------------------------

Jeśli będę chciał zastosować regułę De’Hospitala to MUSZĘ zamienić „n” na „x”!!!! Policzyć granicę, a następnie napisać zdanie:

„Ponieważ granica funkcji …….(ta z „x”) jest równa …….. przy x dążącym do ……(„x” dąży do tego co „n”) , to granica ciągu …….(tego z „n”) jest równa …… ”

NIE WOLNO REGUŁY DE’HOSPITALA STOSOWAĆ PRZY „n” ZAWSZE ZAMIENIAMY NA „x”!!!

Przykład:

$$\operatorname{}\frac{\sin\frac{1}{n}}{\frac{1}{n}} =$$

$$\operatorname{}\frac{\sin\frac{1}{x}}{\frac{1}{x}} = \left\{ \left. \ t = \frac{1}{x} \right\} \right.\ = \operatorname{}\frac{\text{sint}}{t} = \frac{\mathrm{o}}{\mathrm{o}} = \operatorname{}\frac{(sint)'}{t'} = \operatorname{}\text{cost} = 1$$

Ponieważ granica funkcji $\frac{\sin\frac{1}{x}}{\frac{1}{x}}$ jest równa 1 przy × → ∞ , to granica ciągu $\frac{\sin\frac{1}{n}}{\frac{1}{n}}$ jest równa 1.

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---------------------------------------POCHODNA Z DEFINICJI:---------------------------------------

$$f^{'}\left( x \right) = \operatorname{}\frac{f\left( x + h \right) - f(x)}{h}$$

Przykład 1:

$$\left( \text{lnx} \right)^{'} = \operatorname{}\frac{\ln\left( x + h \right) - lnx}{h} = \operatorname{}{\frac{ln(\frac{x + h}{x})}{h} =}\operatorname{}\frac{1}{h}\ln{\left( 1 + \frac{h}{x} \right) =}\operatorname{}{\ln{(1 + \frac{h}{x})}^{\frac{1}{}}} = {\ln{\operatorname{}{(1 + \frac{h}{x}}})}^{\frac{1}{h}} = \ln{\operatorname{}{{(1 + \frac{1}{\frac{x}{h}})}^{\frac{x}{h}*\frac{1}{x}} = \ln{e^{\frac{1}{x}} = e^{\frac{1}{x}}}}}$$

Przykład 2:

$$\left( \sqrt{x} \right)^{'} = \operatorname{}\frac{\sqrt{x + h} - \sqrt{x}}{h} = \operatorname{}\frac{\left( \sqrt{x + h} - \sqrt{x} \right)\left( \sqrt{x + h} + \sqrt{x} \right)}{h\left( \sqrt{x + h} + \sqrt{x} \right)} = \operatorname{}\frac{x + h - x}{h\left( \sqrt{x + h} + \sqrt{x} \right)} = \operatorname{}{\frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}}$$

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