What is the derivative of f(x)=2e^{2e^x+x} ?

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

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

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

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derivative of

f (x) = 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)

\frac{3}{y ^{4}}

y^{^{''}} + 2 y^{^{'}} + y = 2^{- t}

\frac{cos ( x )}{x}

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

\frac{d}{d x} (25 + (12 - x)^{2} + x)

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