What is the first derivative of 2x^2-4 ?

The first derivative of 2x^2-4 is 4x

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derivative of 2x^2-4

\frac{d}{d x} (2 x^{2} - 4)

4 x

Solution steps

\frac{d}{d x} (2 x^{2} - 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{2}) - \frac{d}{d x} (4)

\frac{d}{d x} (2 x^{2}) = 4 x

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

= 4 x - 0

Simplify

= 4 x

\frac{d}{d x} (ln x^{2} - 4)

f (x) = - 2 x^{2} - 12 x - 4

\frac{t d y}{d t} = t^{3} + 8 t^{3} y

\frac{d}{d x} (\frac{( x + 3 ) ^{2}}{e ^{x} \cdot cos ( x )})

f (x) = \frac{1}{5} (9 x + 8)^{3}