If y = x^n x + x ( x)^n , then dy dx = - Vidyadip

MCQ

If $y = {x^n}\log x + x{(\log x)^n}$, then ${{dy} \over {dx}} = $

✓
${x^{n - 1}}(1 + n\log x) + {(\log x)^{n - 1}}[n + \log x]$
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 these

✓

Answer

Correct option: A.

${x^{n - 1}}(1 + n\log x) + {(\log x)^{n - 1}}[n + \log x]$

a
(a) $y = {x^n}\log x + x{(\log x)^n}$

$\frac{{dy}}{{dx}} = n{x^{n - 1}}\log x + {x^n}.\left( {\frac{1}{x}} \right) + xn{(\log x)^{n - 1}}.\left( {\frac{1}{x}} \right) + 1.{(\log x)^n}$

$ = {x^{n - 1}}(1 + n\log x) + {(\log x)^{n - 1}}[n + \log x]$.

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