Derivation y = x / (3x-2) 2 (the last 2 is square)

Derivation y = x / (3x-2) 2 (the last 2 is square)

y=x(3x-2)^(-2)
y'=x'*(3x-2)^(-2)+x*[(3x-2)^(-2)]'
=(3x-2)^(-2)+x*(-2)*(3x-2)^(-3)*(3x-2)'
=(3x-2)^(-2)-6x(3x-2)^(-3)
=(-3x-2)/(3x-2)³

Finding monotone interval y = (x-1) square × (X-2) cube PS: using derivation

y'=2(x-1)(x-2)^3+(x-1)^2*3(x-2)^2=(x-1)(x-2)^2[2(x-2)+3(x-1)]=(x-1)(x-2)^2(5x-7)
When y '> - 0, there is: X7 / 5
y'

Y = cos (2x + 3) + e ^ (- 5x) is the answer - 2Sin (2x + 3-5e ^ (- 5x)?
Finding y '(x) and Dy

Y = cos (2x + 3) + e ^ (- 5x) derivation
dy/dx=-2sin(2x+3)-5e^(-5x),dy=[-2sin(2x+3)-5e^(-5x)]dx.
You're right!

Y = (2x ^ 2-5x + 1) e ^ x,

y'=(2x^2-5x+1) ' e^x+(2x^2-5x+1)(e^x)'
=(4x-5)e^x+(2x^2-5x+1)e^x
=(2x²-x-4)e^x