What is the derivative of y=x^3-3x+1 ?

The derivative of y=x^3-3x+1 is 3x^2-3

What is the first derivative of y=x^3-3x+1 ?

The first derivative of y=x^3-3x+1 is 3x^2-3

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derivative of

y = x^{3} - 3 x + 1

3 x^{2} - 3

Solution steps

\frac{d}{d x} (x^{3} - 3 x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) - \frac{d}{d x} (3 x) + \frac{d}{d x} (1)

\frac{d}{d x} (x^{3}) = 3 x^{2}

\frac{d}{d x} (3 x) = 3

\frac{d}{d x} (1) = 0

= 3 x^{2} - 3 + 0

Simplify

= 3 x^{2} - 3

\frac{\partial}{\partial x} (sin (x) + x^{2} e^{y} - 1)

f (x) = \frac{ln ( x )}{x ^{3}}

x \to 0.01 lim (a x^{2} - x + b)

\int cos (x) (9 + 5 sin^{2} (x)) d x

\int sin (x y) + x y cos (x y) d x

What is the derivative of y=x^3-3x+1 ?
The derivative of y=x^3-3x+1 is 3x^2-3
What is the first derivative of y=x^3-3x+1 ?
The first derivative of y=x^3-3x+1 is 3x^2-3