y=22x ⇒ logy=2xlog2 ⇒ ddxlogy=log2ddx(2x) ⇒1ydydx=log2 2xlog2 ⇒dydx=y(log2)22x

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If $y = 2^{2^{x}}$ then $\frac{d y}{d x} =$

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$y . {(\log 2)}^{2}) 2^{x}$

$y (\log 2) {.2}^{x}$

$y 2 {(\log 2)}^{2}) 2^{x}$

$- y (\log 2) {.2}^{x}$

answer is A.

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Detailed Solution

$y = 2^{2^{x}} \Rightarrow \log y = 2^{x} \log 2 \Rightarrow \frac{d}{d x} \log y = \log 2 \frac{d}{d x} (2^{x}) \Rightarrow \frac{1}{y} \frac{d y}{d x} = \log 2 2^{x} \log 2 \Rightarrow \frac{d y}{d x} = y {(\log 2)}^{2} 2^{x}$

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