Производна на функция

[Гост]

[tex]y = \frac{\arcsin(3x)}{x^2} + 5\ln^3(x) - e^{\sqrt{5x^2}} + \left(\arctan(x)\right)^{\cos(2x)}[/tex]

[ammornil]

[tex]y = \frac{\arcsin(3x)}{x^2} + 5\ln^3(x) - e^{\sqrt{5x^2}} + \left(\arctan(x)\right)^{\cos(2x)}[/tex]

$\\[6pt] \text{ДМ}\quad \begin{array}{|l} -1\le{3x}\le{1} \\ x>0 \\ -1 \le{2x} \le{1} \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} x \in\left[-\dfrac{1}{3};\dfrac{1}{3}\right] \\ x\in(0; +\infty) \\ x\in\left[-\dfrac{1}{2};\dfrac{1}{2}\right] \end{array} \Rightarrow x\in\left(0;\dfrac{1}{3}\right] \\[12pt] \quad \begin{pmatrix}\arcsin{(3x)}\end{pmatrix}'= \dfrac{1}{\sqrt{1-9x^{2}}}\cdot{3}= \dfrac{3}{\sqrt{1-9x^{2}}}, \quad \begin{pmatrix}x^{2}\end{pmatrix}'= 2x \\[6pt] \begin{pmatrix} \dfrac{\arcsin{(3x)}}{x^{2}} \end{pmatrix}'= \dfrac{\dfrac{3}{\sqrt{1-9x^{2}}}\cdot{x^{2}}-2x\cdot{\arcsin{(3x)}}}{x^{4}}=\dfrac{3x-2\sqrt{1-9x^{2}}\arcsin{(3x)}}{x^{3}\sqrt{1-9x^{2}}} \\[6pt] \begin{pmatrix}5\ln^{3}{(x)}\end{pmatrix}'= 3\cdot{5\ln^{2}{(x)}}\cdot{\dfrac{1}{x}}= \dfrac{15\ln^{2}{(x)}}{x} \\[6pt] \begin{pmatrix}e^{\sqrt{5x^{2}}}\end{pmatrix}'= 2\cdot{5x}\cdot{\dfrac{1}{2\sqrt{5x^{2}}}}\cdot{e^{\sqrt{5x^{2}}}}= \sqrt{5}e^{\sqrt{5x^{2}}} \\[6pt] \begin{pmatrix}\arctg^{\cos{(2x)}}{(x)}\end{pmatrix}'= 2\cdot{(-\sin{(2x)})}\cdot{\cos{(2x)}}\arctg^{\cos{(2x)-1}}{(x)}\cdot{\dfrac{1}{1+x^{2}}}=-\dfrac{2\sin{(4x)}}{1+x^{2}}\arctg^{\cos{(2x)-1}}{(x)} \\[12pt] y'= \begin{pmatrix}\dfrac{\arcsin(3x)}{x^2}\end{pmatrix}' + \begin{pmatrix}5\ln^3(x)\end{pmatrix}' - \begin{pmatrix}e^{\sqrt{5x^2}}\end{pmatrix}' + \begin{pmatrix}\arctg^{\cos(2x)}{(x)}\end{pmatrix}'= \cdots\\[24pt]$Прегледайте преобразуванията защото работих директно в LaTeX.