Question

If \(y = cos^{-1}(\frac{1-4^x}{1+4^x})\), then \(\frac{dy}{dx}\) at \(x = 1\) is

• A. \(\frac{4log2}{5}\)
• B. \(\frac{log2}{5}\)
• C. \(\frac{2log8}{5}\)
• D. \(\frac{5log8}{2}\)

Hint:

Write 4^x as (2^x)^2 and use cos(2 theta) form to simplify before differentiating.

Answer 1

The Correct Option is A

Step 1: Direct differentiation
$\frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}}u'$, where $u = \frac{1-4^x}{1+4^x}$.

Step 2: At x = 1
$u = -\frac35$, so $\sqrt{1-u^2} = \frac45$. And $u' = \frac{-4^x\ln4(1+4^x) - (1-4^x)4^x\ln4}{(1+4^x)^2} = \frac{-2\cdot4^x\ln4}{(1+4^x)^2} = \frac{-8\cdot2\ln2}{25}$.

Step 3: Combine
$\frac{dy}{dx} = -\frac54\times\frac{-16\ln2}{25} = \frac{4\ln2}{5}$. Option (A).

Final Answer:
4 log 2 / 5. \[ \boxed{\text{(A)}\ \frac{4\log2}{5}} \]