differentation the last ones?

Question

(x^3 -2x^2 + x-3)/x

(2x-1)^2/x^2


(2x^2-x-1)/x^3

Answer 1

y=(x^3-2x^2+x-3)/x
dy/dx= use the quotient rule
[dr(derivative of Nr) - Nr (Derivaive of Dr)] /Dr^2
Dr=denominator
Nr=Numerator
[x(3x^2-4x+1) - (x^3-2x^2+x-3)(1)] / x^2
=(3x^3-4x^2+x -x^3+2x^2-x+3) /x^2
=(2x^3-2x^2+3)/x^2

use the same principle as before
derivative of Nr= 2(2x-1)
derivative of Dr = 2x

derivative of Nr=(4x-1)
derivative of Dr=3x^2

Answer 2

1)
(x^3 - 2x^2 + x - 3)/x

=>x^2 - 2x + 1 - (3/x)

differentiating

2x - 2 -3(-x^-2)

=>2x - 2 + 3/x^2

=>(2x^3 - 2x^2 + 3)/x^2

2)

(2x -1)^2/x^2

=>(4x^2 - 4x + 1)/x^2

=>4 - (4/x) + (1/x^2)

differentiating

4x +( 4/x^2) - (2/x^3)

=>(4x^4 + 4x -2)/x^3

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

3)

(2x^2 - x - 1)/x^3

differentiating

[x^3(4x - 1) - (2x^2 - x - 1)(3x^2)]/x^6

=>[4x^4 - x^3 - (6x^4 - 3x^3 - 3x^2)]/x^6

=>(4x^4 - x^3 - 6x^4 + 3x^3 + 3x^2)/x^6

=>(-2x^4 + 2x^3 + 3x^2)/x^6

=> (-2x^2 + 2x + 3)/x^4

Answer 3

Simplify first.
f(x) = (x^3 -2x^2 + x-3)/x = x^2 - 2x + 1 - 3/x
f'(x) = 2x - 2 +3/x^2

Can you follow my way to finish the other two?

Answer 4

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

(2x-1)^2/x^2
= [x^2(2x-1)2 -2x(2x-1)^2]/x^4
=[2(2x-1) -2(2x-1)^2]x^3
=[2(2x-1)(1 - 2x-1)}/x^3
= (-4x(2x-1))/x^3

= {x^3(4x-1) -(2x^2-x-1)(3x^2)}/x^6
= [x(4x-1) -3(2x^2-x-1)]/x^4
= [4x^2 -x -6x^2+3x+3]/x^4
= [-2x^2 +2x +3]/x^4


(2x^2-x-1)/x^3