IDZ Ryabushko Option 6 KR "Derivatives and their applic

Kontrol´naya rabota "Proizvodnyye i ikh prilozheniya" (2 chasa). 260 str. 1. Nayti proizvodnuyu pervogo poryadka y′ 2. Nayti proizvodnuyu pervogo poryadka y′ 2.6 y = e−√x/(1 + e2x) 3. Vychislit´ pervuyu proizvodnuyu funktsii pri ukazannom znachenii argumenta ili parametra libo pri zadannykh koordinatakh tochki. 3.6 f(x) = x/(2x – 1), x = –2 4. Nayti vtoruyu proizvodnuyu y′′ 4.6 y = 3√(1 − x)2 5. Nayti vtoruyu proizvodnuyu d2y/dx2 funktsii. 6. Reshit´ sleduyushchiye zadachi. 6.6. Kanat visyachego mosta imeyet formu paraboly i prikreplen k vertikal´nym oporam, otstoyashchim odna ot drugoy na rasstoyanii 200 m. Samaya nizhnyaya tochka kanata nakhoditsya na 40 m nizhe tochek podvesa. Nayti ugol mezhdu kanatom i oporami. 7. Reshit´ sleduyushchiye zadachi. 7.6. Telo brosheno vertikal´no vverkh s nachal´noy skorost´yu a m/s. Za kakoye vremya i na kakom rasstoyanii ot poverkhnosti Zemli telo dostignet naivysshey tochki?
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Test "Derivatives and their applications" (2 hours). 255 pages

1. Find the first derivative of y′

2. Find the first derivative of y′
2.6 y = e−√x/(1 + e2x)

3. Calculate the first derivative of the function for the specified value of the argument or parameter or for the given coordinates of the point.
3.6 f(x) = x/(2x – 1), x = –2

4. Find the second derivative of y′′
4.6 y = 3√(1 − x)2

5. Find the second derivative d2y/dx2 of the function.

6. Solve the following problems.
6.6. The rope of a suspension bridge has the shape of a parabola and is attached to vertical supports spaced 200 m apart. The lowest point of the rope is 40 m below the suspension points. Find the angle between the rope and the supports.

7. Solve the following problems.
7.6. A body is thrown vertically upward with an initial velocity of a m/s. In what time and at what distance from the Earth´s surface will the body reach its highest point?