The correct option is B 2^3x(log_2)^2 Taking log #160; on both side #160; log #160; y= 2^ x log _ 2 #160; and differen ...

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Logarithmic Differentiation

If y=22x, t...

Question

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

y ({log}_{10} 2)^{2}

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y ({log}_{e} 2)^{2}

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2^{3 x} ({log}_{2})^{2}

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y {log}_{e} 2

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Solution

The correct option is B $2^{3 x} ({log}_{2})^{2}$
$\begin{matrix} T a k i n g log o n b o t h s i d e log y = 2^{x} {log}_{2} a n d d i f f e r e n t i a t e w . r . t o x \frac{d (log y)}{d x} = {log}_{2} \frac{d (2^{x})}{d x} \frac{1}{y} . \frac{d y}{d x} = {log}_{2} (2^{x} . {log}_{2}) \frac{d y}{d x} = y {log}_{2} (2^{x} . {log}_{2}) \frac{d y}{d x} = 2^{2^{x}} {({log}_{2})}^{2} {.2}^{x} \frac{d y}{d x} = 2^{3 x} . {({log}_{2})}^{2} \end{matrix}$

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Similar questions

\frac{d y}{d x} - y {log}_{e} 2 = 2^{sin x} (cos x - 1) {log}_{e} 2

y (\frac{π}{2}) = 2,

\frac{d y}{d x} = \frac{2^{x} y + 2^{y} \cdot 2^{x}}{2^{x} + 2^{x + y} {log}_{e}^{2}},

y (0) = 0,

y = 1

x

y = {tan}^{- 1} (\frac{2^{x^{'}}}{1 + 2^{2 x + 1}})

\frac{d y}{d x}

x = 0

If $y = {tan}^{- 1} (\frac{2^{x}}{1 + 2^{2 x + 1}})$ , then $\frac{d y}{d x}$ at $x = 0$ is

y (x)

\frac{d y}{d x} + (\frac{2 x + 1}{x}) y = e^{- 2 x}, x > 0

y (1) = \frac{1}{2} e^{- 2}

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