KCET · Maths · Differentiation
If \( f(x)=\sin ^{-1}\left[\frac{2^{x+1}}{1+4^{x}}\right] \), then \( f^{\prime}(0)= \)
- A \( \log 2 \)
- B \( \frac{4 \log 2}{5} \)
- C \( 2 \log 2 \)
- D \( \frac{2 \log 2}{5} \)
Answer & Solution
Correct Answer
(A) \( \log 2 \)
Step-by-step Solution
Detailed explanation
(A)
\( f(x)=\sin ^{-1}\left[\frac{2 \cdot 2^{x}}{1+\left(2^{x}\right)^{2}}\right] \)
Put \( 2 x=\tan \theta \)
\[
\begin{array}{l}
f(x)=\sin ^{-1}\left[\frac{2 \tan \theta}{1+\left(\tan ^{2} \theta\right)}\right] \\
=\sin ^{-1} \sin 2 \theta \\
f(x)=2 \theta \\
f(x)=2 \tan ^{-1}\left(2^{x}\right) \\
f^{\prime}(x)=\frac{2}{1+\left(2^{x}\right)^{2}} \cdot 2^{x} \log _{e}^{2} \\
f^{\prime}(0)=\log _{e} 2
\end{array}
\]
\( f(x)=\sin ^{-1}\left[\frac{2 \cdot 2^{x}}{1+\left(2^{x}\right)^{2}}\right] \)
Put \( 2 x=\tan \theta \)
\[
\begin{array}{l}
f(x)=\sin ^{-1}\left[\frac{2 \tan \theta}{1+\left(\tan ^{2} \theta\right)}\right] \\
=\sin ^{-1} \sin 2 \theta \\
f(x)=2 \theta \\
f(x)=2 \tan ^{-1}\left(2^{x}\right) \\
f^{\prime}(x)=\frac{2}{1+\left(2^{x}\right)^{2}} \cdot 2^{x} \log _{e}^{2} \\
f^{\prime}(0)=\log _{e} 2
\end{array}
\]
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