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e^x*log(sin(2*x), 10) |
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Answer» 2ex tan 2x + ex log con2x is the correct answer y=e^x log(sin2x)lny=xlog(sin2x)log(sin2x) =z10^z=sin(2x)or z ln10=lnsin(2x)i.e z =lnsin(2x)/ln10Hence lny =xln(sin2x)/ln10Differentiating both sides wrt 'x'we get(1/y)(dy/dx)=[lnsin(2x)+x{2cos(2x)/sin2x}]/ln10or dy/dx= y[lnsin(2x)+x{2cos(2x)/sin2x}]/ln10={xlog(sin2x)}×[lnsin(2x)+x{2cos(2x)/sin2x}]/ln10 |
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