What is the first (1+e^{2x})^' ?

The first (1+e^{2x})^' is 2e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1+e^{2x})^'

(1 + e^{2 x})^{^{'}}

2 e^{2 x}

Solution steps

(1 + e^{2 x})^{'}

Apply the Sum/Difference Rule:

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

= 1^{^{'}} + (e^{2 x})^{^{'}}

1^{'} = 0

(e^{2 x})^{'} = e^{2 x} \cdot 2

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

Simplify

= 2 e^{2 x}

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

x \to \frac{2 π}{3} lim (- tan (2 x))

y^{^{'}} + \frac{2}{200 + t} y = 3

(x - 4) (x + 5) (x + 4)