First slide

Derivatives

Question

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

Moderate

A

1
B

2
C

log 2
D

$- \frac{1}{10} \log 2$

Solution

$y = \tan^{- 1} (\frac{2^{x + 1} - 2^{x}}{1 + 2^{x} {.2}^{x + 1}})$
$\begin{array}{l} = {Tan}^{- 1} \cdot 2^{x + 1} - {Tan}^{- 1} 2^{x} \\ y^{'} = \frac{2^{x + 1} \log 2}{1 + {(2^{x + 1})}^{2}} - \frac{2^{x} \log 2}{1 + {(2^{x})}^{2}} \\ y^{'} (0) = - \frac{1}{10} \log 2 \end{array}$

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