KCET · Maths · Differentiation
If \( y=\log (\log x) \) then \( \frac{d^{2} y}{d x^{2}} \) is equal to
- A \( -\frac{(1+\log x)}{(x \log x)^{2}} \)
- B \( -\frac{(1+\log x)}{x^{2} \log x} \)
- C \( \frac{(1+\log x)}{(x \log x)^{2}} \)
- D \( \frac{(1+\log x)}{x^{2} \log x} \)
Answer & Solution
Correct Answer
(A) \( -\frac{(1+\log x)}{(x \log x)^{2}} \)
Step-by-step Solution
Detailed explanation
Given that, \( y=\log (\log x) \rightarrow(1) \)
Differentiate Eq. (1) with respect to \( x \), we get
\[
\frac{d y}{d x}=\frac{1}{\log x}\left(\frac{1}{x}\right)=\frac{1}{x \log x}
\]
Again, differentiate Eq. (2) with respect to \( x \), we get
\[
\begin{array}{l}
\frac{d^{2} y}{d x}=\frac{-1}{(x \log x)^{2}} \cdot\left[x \cdot \frac{1}{x}+\log x \cdot 1\right] \\
=\frac{-(1+\log x)}{(x \log x)^{2}}
\end{array}
\]
Differentiate Eq. (1) with respect to \( x \), we get
\[
\frac{d y}{d x}=\frac{1}{\log x}\left(\frac{1}{x}\right)=\frac{1}{x \log x}
\]
Again, differentiate Eq. (2) with respect to \( x \), we get
\[
\begin{array}{l}
\frac{d^{2} y}{d x}=\frac{-1}{(x \log x)^{2}} \cdot\left[x \cdot \frac{1}{x}+\log x \cdot 1\right] \\
=\frac{-(1+\log x)}{(x \log x)^{2}}
\end{array}
\]
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