Đáp án:
 
Giải thích các bước giải:
 

@Sun 
y = 1/3x^6 + 3x^2 + 2
y' = 2x^5 + 6x  
b) y = (3x+5)/(1-2x) 
y' = [(3x+5)'(1-2x) - (1-2x)'(3x+5)]/[(1-2x)^2] 
y' = [3(1-2x) + 2(3x+5)]/[(1-2x)^2] 
y' = (3-6x + 6x + 10)/[(1-2x)^2] 
y' =13/[(1-2x)^2] 
c) y = ( sqrt(3x+5) + 1)^8 
y' = 8 . (sqrt(3x+5) + 1)' . (sqrt(3x+5) + 1)^7 
y' = 8. [(3x+5)']/[2sqrt(3x+5)] .(sqrt(3x+5) +1)^7 
y' = 8. 3/[2sqrt(3x+5)] .(sqrt(3x+5) +1)^7
y' = [21 .(sqrt(3x+5)+1)^7]/[2sqrt(3x+5)]
d) y = (3x+4)/(sqrt(1-3x)) 
y' = [(3x+4)'sqrt(1-3x) - (sqrt(1-3x))'(3x+4)]/(1-3x) 
y' = [3sqrt(1-3x) -(-3)/[2sqrt(1-3x)] (3x+4)]/(1-3x) 
y' = [ (6(1-3x)+ 3(3x+4))/(2sqrt(1-3x)) ]/(1-3x) 
y' = (6 -18x + 9x + 12)/[2sqrt(1-3x)(1-3x)] 
y' = (-9x +18)/[2sqrt(1-3x)(1-3x)]