Right choice is (d) x^3e^3x.3e^3x (3 logx+\(\frac{1}{x}\))
For explanation: Consider y=x^3e^3x
Applying log on both sides, we get
logy=3e^3x logx
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3e^{3x}) logx+\frac{d}{dx} (logx)3e^{3x}\)
\(\frac{1}{y} \frac{dy}{dx}\)=3e^3x.3.logx+\(\frac{1}{x}\) 3e^3x
\(\frac{dy}{dx}\)=y(3e^3x.3.logx+\(\frac{1}{x}\) 3e^3x)
\(\frac{dy}{dx}\)=x^3e^3x.3e^3x (3 logx+\(\frac{1}{x}\))