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प्रश्न
xn loga x
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उत्तर
\[\text{ Let } u = x^n ; v = \log_a x = \frac{\log x}{\log a}\]
\[\text{ Then }, u' = n x^{n - 1} ; v' = \frac{1}{x \log a}\]
\[\text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uv \right) = uv' + vu'\]
\[\frac{d}{dx}\left( x^n \log_a x \right) = x^n . \frac{1}{x \log a} + \log_a x \left( n x^{n - 1} \right)\]
\[ = x^{n - 1} \frac{1}{\log a} + \log_a x \left( n x^{n - 1} \right)\]
\[ = x^{n - 1} \left( \frac{1}{\log a} + n \log_a x \right)\]
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