Question

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

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

Answer 1

Answer: (B)

$\frac{1}{10}$log_e2

Answer 2

y = tan^–1 $\left( \frac{2^{x}}{1 + 2^{x + 1}} \right)$

$\frac{dy}{dx}$= $\frac{1}{1 + \left( \frac{2^{x}}{1 + 2^{x + 1}} \right)^{2}}$ × $\frac{(1 + 2^{x + 1})2^{x}\log 2 - 2^{x}.2^{x + 1}\log 2}{(1 + 2^{x + 1})^{2}}$r = 0

= $\frac{1}{1 + \left( \frac{1}{3} \right)^{2}}$. $\frac{3.\log 2 - 1.2\log 2}{9}$

= $\frac{1}{10}$. $\frac{(\log 2^{3} - \log 2^{2})}{1}$= $\frac{1}{10}$log 2