What is the derivative of 2e^{3x}+1 ?

The derivative of 2e^{3x}+1 is 6e^{3x}

What is the first derivative of 2e^{3x}+1 ?

The first derivative of 2e^{3x}+1 is 6e^{3x}

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\frac{d}{d x} (2 e^{3 x} + 1)

6 e^{3 x}

Solution steps

\frac{d}{d x} (2 e^{3 x} + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 e^{3 x}) + \frac{d}{d x} (1)

\frac{d}{d x} (2 e^{3 x}) = 6 e^{3 x}

\frac{d}{d x} (1) = 0

= 6 e^{3 x} + 0

Simplify

= 6 e^{3 x}

\int sec (π x) tan (π x) d x

\int 8 e^{8 x - 3} d x

x \to 6 lim ((\frac{1}{x - 6})^{2})

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

y = \frac{25 - x ^{2}}{4 x}

What is the derivative of 2e^{3x}+1 ?
The derivative of 2e^{3x}+1 is 6e^{3x}
What is the first derivative of 2e^{3x}+1 ?
The first derivative of 2e^{3x}+1 is 6e^{3x}