Сечения и производные сложных функций

Решите задачу:

$(e^{2x})'=e^{2x}\cdot (2x)'=2e^{2x}\\\\(sin3x)'=-cos3x\cdot (3x)'=-3cos3x\\\\(arcsin6x)'=\frac{1}{\sqrt{1-(6x)^2}}\cdot (6x)'=\frac{6}{\sqrt{1-36x^2}}\\\\(x^3+x^2-0,5)'=3x^2+2x-0=3x^2+2x\\\\(e^{3x+1})'=e^{3x+1}\cdot (3x+1)'=3e^{3x+1}\\\\(ctgx^2)'=-\frac{1}{sin^2x^2}\cdot (x^2)'=-\frac{2x}{sin^2x^2}\\\\(\sqrt{12x})'=\frac{1}{2\sqrt{12x}}\cdot (12x)'=\frac{12}{2\sqrt{12x}}=\frac{6}{\sqrt{12x}}\\\\(tg\frac{x}{2})'=\frac{1}{cos^2\frac{x}{2}}\cdot (\frac{x}{2})'=\frac{1}{2cos^2\frac{x}{2}}\\\\(3\sqrt{x})'=3\cdot \frac{1}{2\sqrt{x}}$

$(\frac{1}{x^2})'=(x^{-2})'=-2\cdot x^{-3}=-\frac{2}{x^3}\\\\(\frac{1}{cosx})'=-\frac{-sinx}{cos^2x}=\frac{sinx}{cos^2x}\; \quad [\, (\frac{1}{u})'=-\frac{u'}{u^2}\, ]$