MHT CET · Maths · Differentiation
The derivative of \((\log x)^{x}\) with respect to \(\log x\) is
- A \((\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
- B \((\log x)^{x}\left[\log x+\frac{1}{\log (\log x)}\right]\)
- C \(x(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
- D None of the above
Answer & Solution
Correct Answer
(C) \(x(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
Step-by-step Solution
Detailed explanation
Let \(u=(\log x)^{x}\)
\(\Rightarrow \log u=x \log (\log x) \)
\( \therefore \frac{1}{u} \frac{d u}{d x}=x \frac{1}{\log x} \cdot \frac{1}{x}+\log (\log x) \)
\( \Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
and \(v=\log x\)
\(\therefore \frac{d v}{d x}=\frac{1}{x}\)
Now, \(\frac{d u}{d v}=(\log x)^{x}\left[\frac{1}{\log x} \times \log (\log x)\right] \times x\)
\(=x(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
\(\Rightarrow \log u=x \log (\log x) \)
\( \therefore \frac{1}{u} \frac{d u}{d x}=x \frac{1}{\log x} \cdot \frac{1}{x}+\log (\log x) \)
\( \Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
and \(v=\log x\)
\(\therefore \frac{d v}{d x}=\frac{1}{x}\)
Now, \(\frac{d u}{d v}=(\log x)^{x}\left[\frac{1}{\log x} \times \log (\log x)\right] \times x\)
\(=x(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]\)
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