J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    If \[y=6\,\,\sin x\,{{\log }_{10}}x+{{e}^{{{x}^{2}}}},\] then \[\frac{dy}{dx}\] is equal to

    A)  \[\frac{6\,\,\sin x}{x}+6\,\cos \,x\,{{\log }_{10}}\,x+2x{{e}^{{{x}^{2}}}}\]

    B)  \[\frac{6\,\,\sin x}{x}\,{{\log }_{10}}\,e+6\,\cos \,x\,{{\log }_{10}}x+2x{{e}^{{{x}^{2}}}}\]

    C)  \[6\,\cos \,x\,{{\log }_{10}}x+6\,\sin x.\frac{10}{x}+2x{{e}^{{{x}^{2}}}}\]

    D)  \[\frac{6\,\sin \,x}{x}+6\,\cos \,x\,{{\log }_{10}}x+x{{e}^{{{x}^{2}}}}\]

    Correct Answer: B

    Solution :

    \[y=6\,\sin x\,{{\log }_{10}}\,x+{{e}^{{{x}^{2}}}}\] \[\frac{dy}{dx}=6\left\{ \sin x.\frac{1}{x}.{{\log }_{10}}e+\cos x.{{\log }_{10}}x \right\}\] \[+{{e}^{{{x}^{2}}}}\,.2x\] \[\frac{dy}{dx}=\frac{6\,\sin x}{x}.{{\log }_{10}}e+6\,\cos x.{{\log }_{10}}x+2x{{e}^{{{x}^{2}}}}\]


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