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

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

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

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

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derivative of

y = x^{3} - 3 x

3 x^{2} - 3

Solution steps

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

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

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

= 3 x^{2} - 3

\frac{d}{d x} (\frac{1 + sin ( x )}{1 + cos ( x )})

(2 r a^{r x} + n)^{p}

x \to 0 + lim (x^{x^{x}})

\frac{d}{d x} (x - 0.8 - 0.2 sin (x))

n = 0 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

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