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

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

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

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

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

2 e^{2 x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= - 0 + e^{2 x} \cdot 2

Simplify

= 2 e^{2 x}

y = e^{x}, y = e^{3 x}, x = 0, x = ln (4)

\int_{0}^{1} 7 cos (4 x) d x

\int \frac{2}{x} - \frac{3}{x ^{4}} d x

\frac{d}{d x} y + 3 y = 0

(5, - 3), (- 1, - 1)

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