Question

If $y = {\log _{{x^2} + 4}}\left( {7{x^2} - 5x + 1} \right)$, then $\frac{{dy}}{{dx}}$ is equal to

Accepted Answer

$\frac{1}{{{{\log }_e}\left( {{x^2} + 4} \right)}}\left\{ {\frac{{14x - 5}}{{7{x^2} - 5x + 1}} - \frac{{2xy}}{{{x^2} + 4}}} \right\}$

Answer

${\log _e}\left( {{x^2} + 4} \right).\left\{ {\frac{{14x - 5}}{{7{x^2} - 5x + 1}} - \frac{{2xy}}{{{x^2} + 4}}} \right\}$

Answer

${\log _e}\left( {7{x^2} - 5x + 1} \right)\left\{ {\frac{{2x}}{{{x^2} + 4}} - \frac{{\left( {14x - 5} \right)y}}{{7{x^2} - 5x + 1}}} \right\}$

Answer

$\frac{1}{{{{\log }_e}\left( {7{x^2} - 5x + 1} \right)}}\left\{ {\frac{{2x}}{{{x^2} + 4}} - \frac{{\left( {14x - 5} \right)y}}{{7{x^2} - 5x + 1}}} \right\}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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