MHT CET · Maths · Differentiation
If \(y=x^{n} \log x+x(\log x)^{n}\), then \(\frac{d y}{d x}\) is equal to
- A \(x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]\)
- B \(x^{n-2}(1+n \log x)+(\log x)^{n-1}[n+\log x]\)
- C \(x^{n-1}(1+n \log x)+(\log x)^{n-1}[n-\log x]\)
- D None of the above
Answer & Solution
Correct Answer
(A) \(x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]\)
Step-by-step Solution
Detailed explanation
Given, \(y=x^{n} \log x+x(\log x)^{n}\)
\(\frac{d y}{d x}=n x^{n-1} \log x+x^{n} \cdot \frac{1}{x}+x n(\log x)^{n-1}\left(\frac{1}{x}\right) +1 \cdot(\log x)^{n} \)
\( =x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]\)
\(\frac{d y}{d x}=n x^{n-1} \log x+x^{n} \cdot \frac{1}{x}+x n(\log x)^{n-1}\left(\frac{1}{x}\right) +1 \cdot(\log x)^{n} \)
\( =x^{n-1}(1+n \log x)+(\log x)^{n-1}[n+\log x]\)
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