What is the (x^3-6x^2+2x-2)^' ?

The (x^3-6x^2+2x-2)^' is 3x^2-12x+2

What is the first (x^3-6x^2+2x-2)^' ?

The first (x^3-6x^2+2x-2)^' is 3x^2-12x+2

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(x^{3} - 6 x^{2} + 2 x - 2)^{^{'}}

3 x^{2} - 12 x + 2

Solution steps

(x^{3} - 6 x^{2} + 2 x - 2)^{'}

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (x^{3})^{^{'}} - (6 x^{2})^{^{'}} + (2 x)^{^{'}} - 2^{^{'}}

(x^{3})^{'} = 3 x^{2}

(6 x^{2})^{'} = 12 x

(2 x)^{'} = 2

2^{'} = 0

= 3 x^{2} - 12 x + 2 - 0

Simplify

= 3 x^{2} - 12 x + 2

sin (x) + \frac{1}{2} cot (x)

y = \frac{ln ( x )}{1 + ln ( x )}

\frac{d}{d x} ((x^{3} + 5) (5 x^{2} - 5))

x y^{2} \frac{d y}{d x} = y^{3} - x^{3}, y (1) = 4

cos (3 x^{2})