Question

If y = tan^–1$\left( \frac{2^{x}}{1 + 2^{x + 1}} \right)$then $\frac{dy}{dx}$ at x = 0 is

• A. $–\frac{3}{5}$log_e2
• B. $\frac{1}{10}$ log_e2
• C. 2
• D. None

Answer 1

Answer: (B)

$\frac{1}{10}$ log_e2

Answer 2

$\frac{dy}{dx} = \frac{1}{1 + \left( \frac{2^{x}}{1 + 2^{(x + 1)}} \right)^{2}} \times \frac{(1 + 2^{x + 1}).2^{x}\log 2 - 2^{x}.2^{x + 1}\log 2}{(1 + 2^{x + 1})^{2}}$

$\frac{dy}{dx}_{(x = 0)} = \frac{1}{10}$ log_e2