What is the derivative of y=2e^{2e^x+x} ?

The derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)

What is the first derivative of y=2e^{2e^x+x} ?

The first derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

y = 2 e^{2 e^{x} + x}

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

What is the derivative of y=2e^{2e^x+x} ?
The derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)
What is the first derivative of y=2e^{2e^x+x} ?
The first derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)