Question

f(x)=(2-3x)/(x-1)-x^4*корень из(2-x) +3sin(пи/2-2x)

Answer 1

F(x)=(2-3x)/(x-1)-x^4*корень из(2-x) +3sin(пи/2-2x)
F'(x)=[(2-3x)/(x-1)]'+[x^4*√(2-x)]'+[3sin(пи/2-2x)]'
[(2-3x)/(x-1)]'=[u/v]'=1/v²[u'v-v'u]   u=2-3x  u'=-3   v=x-1   v'=1
[(2-3x)/(x-1)]'=1/x²[-3x²+3-2-3x]=-1/x²[3x²+3x-1]

[x^4*√(2-x)]'=u'v+v'u   u=x⁴   u'=4x³  v=√(2-x)  v'=1/2√(2-x) 
[x^4*√(2-x)]'=4x³*√(2-x)  +  x⁴/2√(2-x)  

[3sin(пи/2-2x)]'=3*(-2)cos(пи/2-2x)=-6sin2x

F'(x)=-1/x²[3x²+3x-1]+4x³*√(2-x)  +  x⁴/2√(2-x)-6sin2x