Giúp em giải bài này với

`a)`

`y=2x^3-[x^2]/2+4x-1/3`

`-> y'=(2x^3-[x^2]/2+4x-1/3)'`

`-> y'=2(x^3)'-1/2(x^2)'+4x'-(1/3)'`

`-> y'=2*3x^2-1/2*2x+4`

`-> y'=6x^2-x+4`

`b)`

`y=[-2x+3]/[x-4]`

`-> y'=[(-2x+3)'(x-4)-(x-4)'(-2x+3)]/[(x-4)^2]`

`-> y'=[-2(x-4)-1(-2x+3)]/[(x-4)^2]`

`-> y'=[-2x+8+2x-3]/[(x-4)^2]`

`-> y'=5/[(x-4)^2]`

`c)`

`y=[x^2-2x+3]/[x-1]`

`-> y'=[(x^2-2x+3)'(x-1)-(x-1)'(x^2-2x+3)]/[(x-1)^2]`

`->y'=[(2x-2)(x-1)-1(x^2-2x+3)]/[(x-1)^2]`

`-> y'=[2x^2-4x+2-x^2+2x-3]/[(x-1)^2]`

`-> y'=[x^2-2x-1]/[(x-1)^2]`

`d)`

`y=sqrt[5x]`

`-> y'=[(5x)']/[2sqrt[5x]]`

`-> y'=[5]/[2sqrt[5x]]`