Witaj :)
Zad.1. Wykonać działania na macierzach:
a) [tex]C\cdot D[/tex] gdzie:
[tex]C=\left[\begin{array}{ccc}4&5&6\end{array}\right][/tex]
[tex]D=\left[\begin{array}{ccc}1\\2\\3\end{array}\right][/tex]
Mnożenie macierzy przez macierz
[tex]C\cdot D =\left[\begin{array}{ccc}4&5&6\end{array}\right] \cdot \left[\begin{array}{ccc}1\\2\\3\end{array}\right] = \left[\begin{array}{ccc}(4\cdot1+5\cdot2+6\cdot3)\end{array}\right] =\left[\begin{array}{ccc}(4+10+28)\end{array}\right]=\\\\ =\left[\begin{array}{ccc}32\end{array}\right][/tex]
b) [tex]2A+ B^T[/tex] gdzie:
[tex]A=\left[\begin{array}{ccc}-1&0\\-2&1\\3&0\end{array}\right][/tex]
[tex]B=\left[\begin{array}{ccc}2&4&-6\\0&-2&0\end{array}\right][/tex]
Wykonajmy działania osobno
[tex]2A=2\cdot A=2\cdot \left[\begin{array}{ccc}-1&0\\-2&1\\3&0\end{array}\right]=\left[\begin{array}{ccc}2\cdot (-1)&2\cdot 0\\2\cdot (-2)&2\cdot 1\\2\cdot 3&2\cdot 0\end{array}\right] =\left[\begin{array}{ccc}-2&0\\-4&2\\6&0\end{array}\right][/tex]
[tex]B^T=\left[\begin{array}{ccc}2&4&-6\\0&-2&0\end{array}\right]^T=\left[\begin{array}{ccc}2&0\\4&-2\\-6&0\end{array}\right][/tex]
Transponowanie macierzy polega na zamianie wierszy na kolumny lub na odwrót
[tex]2A+ B^T=\left[\begin{array}{ccc}-2&0\\-4&2\\6&0\end{array}\right]+\left[\begin{array}{ccc}2&0\\4&-2\\-6&0\end{array}\right]=\left[\begin{array}{ccc}-2+2&0+0\\-4+4&2+(-2)\\6+(-6)&0+0\end{array}\right] =\left[\begin{array}{ccc}0&0\\0&0\\0&0\end{array}\right][/tex]
Zad.2. Obliczyć całkę nieoznaczoną:
[tex]\int(\sqrt[4]{x} +\frac{3}{x} -\frac{4}{x^2} -5e^x)dx=\int\sqrt[4]{x}\ dx\ +\int\frac{3}{x} dx-\int\frac{4}{x^2} dx-\int5e^xdx=\\\\=\int x^{\frac{1}{4}}dx+3\int\frac{1}{x}dx -4\int\frac{1}{x^2}dx -5\int e^xdx=\\\\=\frac{x^{\frac{5}{4} }}{\frac{5}{4} } +3ln|x|-4(-\frac{1}{x} )-5e^x+C=\\\\=\frac{4\sqrt[4]{x^5} }{5} +3ln|x|+\frac{4}{x}-5e^x+C,\ C\in R[/tex]
Zad.3. Obliczyć pochodną funkcji:
a)
[tex]y=\frac{5x^2}{10x^4-3x^2+7} \\\\y'=(\frac{5x^2}{10x^4-3x^2+7} )'=\frac{(5x^2)'(10x^4-3x^2+7)-5x^2(10x^4-3x^2+7)'}{(10x^4-3x^2+7)^2} =\\\\=\frac{10x(10x^4-3x^2+7)-5x^2(40x^3-6x)}{(10x^4-3x^2+7)^2}=\frac{100x^5-30x^3+70x-200x^5+30x^3}{(10x^4-3x^2+7)^2}=\frac{-100x^5+70x}{(10x^4-3x^2+7)^2} =\\\\=\frac{-10x(10x^4-7)}{(10x^4-3x^2+7)^2}[/tex]
b)
[tex]y= \frac{4}{x} +\sqrt{x} \\\\y'= ( \frac{4}{x} +\sqrt{x})'= (\frac{4}{x} )'+(\sqrt{x} )'=-\frac{4}{x^2}+\frac{1}{2\sqrt{x} }[/tex]
