What is the first (2x^3-3x+1)^' ?

The first (2x^3-3x+1)^' is 6x^2-3

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

(2 x^{3} - 3 x + 1)^{^{'}}

6 x^{2} - 3

Solution steps

(2 x^{3} - 3 x + 1)^{'}

Apply the Sum/Difference Rule:

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

= (2 x^{3})^{^{'}} - (3 x)^{^{'}} + 1^{^{'}}

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

(3 x)^{'} = 3

1^{'} = 0

= 6 x^{2} - 3 + 0

Simplify

= 6 x^{2} - 3

y = (x^{2} - 2 x + 7) 25 - x^{2}

x \to 5 lim (x - 3)

\frac{\partial}{\partial x} (in (x + 2 y + 3 z))

x \to 1 lim (\frac{1}{1 - x})

\int \frac{( ln ( x ) ) ^{48}}{x} d x