What is the first derivative of 2^3x ?

The first derivative of 2^3x is 8

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derivative of 2^3x

derivative of

2^{3} x

8

Solution steps

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

Simplify

2^{3} x : 8 x

= \frac{d}{d x} (8 x)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d x}{d x}

Apply the common derivative:

\frac{d x}{d x} = 1

= 8 \cdot 1

Simplify

= 8

d er i v a t i v e y = x^{2} e^{2}

d er i v a t i v e lo g_{10} (x^{2} + a^{2})

d er i v a t i v e x^{2} sec (x)

d er i v a t i v e x \cdot sin (3 y) + x \cdot e^{\frac{y}{x}}

d er i v a t i v e y = \frac{x + 2}{x ^{2}}