Ответ:
Пошаговое объяснение:
1)
((7x+4)sin8x)' = [(uv)' = uv'+u'v; u=7x-4; v=sin8x] = (7x+4)(sin8x)' + (7x+4)'sin8x
= [ (7x-4)' = 7; (sin8x)' = 8cos8x] = 7sin8x +(7x+4)8cos8x
2)
![\frac{-10x+9}{cos11x} = \left[\begin{array}{ccc}\frac{1}{cosx} = secx\end{array}\right] = (-10x+9)sec11x =](https://tex.z-dn.net/?f=%5Cfrac%7B-10x%2B9%7D%7Bcos11x%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B1%7D%7Bcosx%7D%20%3D%20secx%5Cend%7Barray%7D%5Cright%5D%20%3D%20%28-10x%2B9%29sec11x%20%3D "\frac{-10x+9}{cos11x} = \left[\begin{array}{ccc}\frac{1}{cosx} = secx\end{array}\right] = (-10x+9)sec11x =")
[(uv)' = uv'+u'v; u=-10x+9; v=sec11x] = (-10x+9)' sec11x + (-10x+9)(sec11x)' =
[(-10x+9)'= -10; (sec11x)'= 11(sec11x)(tg11x) ] =-10sec11x +(-10x+9)11(sec11x)(tg11x) =
3)
![5^{-2x+8} ' = \left[\begin{array}{ccc}u=8-2x\end{array}\right] = (5^{u})' = 5^{u} u' =](https://tex.z-dn.net/?f=5%5E%7B-2x%2B8%7D%20%27%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Du%3D8-2x%5Cend%7Barray%7D%5Cright%5D%20%3D%20%285%5E%7Bu%7D%29%27%20%3D%205%5E%7Bu%7D%20u%27%20%3D "5^{-2x+8} ' = \left[\begin{array}{ccc}u=8-2x\end{array}\right] = (5^{u})' = 5^{u} u' =")
