If $y = \log x.{e^{( an x + {x^2})}},$then $\frac{{dy}}{{dx}} = $

Question

If $y = \log x.{e^{(	an x + {x^2})}},$then $\frac{{dy}}{{dx}} = $

Accepted Answer

${e^{(	an x + {x^2})}}\left[ {\frac{1}{x} + ({{\sec }^2}x + 2x)\log x} ight]$

Answer

${e^{(	an x + {x^2})}}\left[ {\frac{1}{x} + ({{\sec }^2}x + x)\log x} ight]$

Answer

${e^{(	an x + {x^2})}}\left[ {\frac{1}{x} + ({{\sec }^2}x - x)\log x} ight]$

Answer

${e^{(	an x + {x^2})}}\left[ {\frac{1}{x} + ({{\sec }^2}x - 2x)\log x} ight]$

differentiation