What is the derivative of 2^{3^{4^x}} ?

The derivative of 2^{3^{4^x}} is ln^2(2)ln(3)*3^{4^x}*2^{3^{4^x}+2x+1}

What is the first derivative of 2^{3^{4^x}} ?

The first derivative of 2^{3^{4^x}} is ln^2(2)ln(3)*3^{4^x}*2^{3^{4^x}+2x+1}

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Popular >

\frac{d}{d x} (2^{3^{4^{x}}})

ln^{2} (2) ln (3) \cdot 3^{4^{x}} \cdot 2^{3^{4^{x}} + 2 x + 1}

Solution steps

\frac{d}{d x} (2^{3^{4^{x}}})

Apply exponent rule:

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

2^{3^{4^{x}}} = e^{3^{4^{x}} l n (2)}

= \frac{d}{d x} (e^{3^{4^{x}} l n (2)})

Apply the chain rule:

e^{3^{4^{x}} l n (2)} \frac{d}{d x} (3^{4^{x}} ln (2))

= e^{3^{4^{x}} l n (2)} \frac{d}{d x} (3^{4^{x}} ln (2))

\frac{d}{d x} (3^{4^{x}} ln (2)) = ln^{2} (2) ln (3) \cdot 3^{4^{x}} \cdot 2^{2 x + 1}

= e^{3^{4^{x}} l n (2)} ln^{2} (2) ln (3) \cdot 3^{4^{x}} \cdot 2^{2 x + 1}

Simplify

e^{3^{4^{x}} l n (2)} ln^{2} (2) ln (3) \cdot 3^{4^{x}} \cdot 2^{2 x + 1} : ln^{2} (2) ln (3) \cdot 3^{4^{x}} \cdot 2^{3^{4^{x}} + 2 x + 1}

= ln^{2} (2) ln (3) \cdot 3^{4^{x}} \cdot 2^{3^{4^{x}} + 2 x + 1}

\int e^{2 x^{2}} x d x

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\int \frac{8}{1 - x ^{2}} d x

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What is the derivative of 2^{3^{4^x}} ?
The derivative of 2^{3^{4^x}} is ln^2(2)ln(3)*3^{4^x}*2^{3^{4^x}+2x+1}
What is the first derivative of 2^{3^{4^x}} ?
The first derivative of 2^{3^{4^x}} is ln^2(2)ln(3)*3^{4^x}*2^{3^{4^x}+2x+1}