What is the derivative of 2^{2^n} ?

The derivative of 2^{2^n} is ln^2(2)*2^{2^n+n}

What is the first derivative of 2^{2^n} ?

The first derivative of 2^{2^n} is ln^2(2)*2^{2^n+n}

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derivative of

2^{2^{n}}

ln^{2} (2) \cdot 2^{2^{n} + n}

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{2^{n}} = e^{2^{n} l n (2)}

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

Apply the chain rule:

e^{2^{n} l n (2)} \frac{d}{d n} (2^{n} ln (2))

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

\frac{d}{d n} (2^{n} ln (2)) = ln^{2} (2) \cdot 2^{n}

= e^{2^{n} l n (2)} ln^{2} (2) \cdot 2^{n}

Simplify

e^{2^{n} l n (2)} ln^{2} (2) \cdot 2^{n} : ln^{2} (2) \cdot 2^{2^{n} + n}

= ln^{2} (2) \cdot 2^{2^{n} + n}

\int \frac{1}{x ^{2} a ^{2} + x ^{2}} d x

e^{- t}

\frac{d}{d x} (c x^{- 4})

\int (2 x^{2} + 9) (x - 3)^{8} d x

\frac{\partial}{\partial x} (x^{2} - y^{2})

What is the derivative of 2^{2^n} ?
The derivative of 2^{2^n} is ln^2(2)*2^{2^n+n}
What is the first derivative of 2^{2^n} ?
The first derivative of 2^{2^n} is ln^2(2)*2^{2^n+n}