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

y^{^{'}} + 5 y = 3

\frac{x ^{4}}{4} + \frac{7}{6} x^{3} - x^{2} + 6 x - 8

x \cdot 3^{x^{2} + 1}

6 (t + 1) (\frac{d y}{d t}) - 5 y = 5 t

x \to \frac{π}{2} lim (cos (3 x))