What is the derivative of 2nde^{2e^x} ?

The derivative of 2nde^{2e^x} is 4dne^{2e^x+x}

What is the first derivative of 2nde^{2e^x} ?

The first derivative of 2nde^{2e^x} is 4dne^{2e^x+x}

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derivative of

2 n d e^{2 e^{x}}

4 d n e^{2 e^{x} + x}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 2 n d e^{2 e^{x}} \cdot 2 e^{x}

Simplify

2 n d e^{2 e^{x}} \cdot 2 e^{x} : 4 d n e^{2 e^{x} + x}

= 4 d n e^{2 e^{x} + x}

n = 1 \sum \infty \frac{n ^{2}}{3}

(3) (6.9)

\frac{d}{d x} ((x + 6)^{2})

x \to 0 lim (\frac{sin ( k ) x}{x})

What is the derivative of 2nde^{2e^x} ?
The derivative of 2nde^{2e^x} is 4dne^{2e^x+x}
What is the first derivative of 2nde^{2e^x} ?
The first derivative of 2nde^{2e^x} is 4dne^{2e^x+x}