What is the derivative of y=x^3-4x+4 ?

The derivative of y=x^3-4x+4 is 3x^2-4

What is the first derivative of y=x^3-4x+4 ?

The first derivative of y=x^3-4x+4 is 3x^2-4

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derivative of

y = x^{3} - 4 x + 4

3 x^{2} - 4

Solution steps

\frac{d}{d x} (x^{3} - 4 x + 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) - \frac{d}{d x} (4 x) + \frac{d}{d x} (4)

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

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (4) = 0

= 3 x^{2} - 4 + 0

Simplify

= 3 x^{2} - 4

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What is the derivative of y=x^3-4x+4 ?
The derivative of y=x^3-4x+4 is 3x^2-4
What is the first derivative of y=x^3-4x+4 ?
The first derivative of y=x^3-4x+4 is 3x^2-4