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?
Ещё
869 / 5 000
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?

Detailed solution. Designed in PDF format for easy viewing of IDZ solutions on smartphones. In MS Word (doc format) sent additionally.