#\frac{d}{dx}((e^x-3x)(\cot (x)+x^2))#
Applying product rule as : #(f\cdot g)^'=f^'\cdot g+f\cdot g^'#
#f=e^x-3x,g=\cot (x)+x^2#
#=\frac{d}{dx}(e^x-3x)(\cot (x)+x^2)+\frac{d}{dx}(\cot \(x)+x^2)(e^x-3x)#............ #eq^n (i)#
Here,
#\frac{d}{dx}(e^x-3x)=e^x-3#
{Applying sum and differnce rule as: #(f\pm g)^'=f^'\pm g^'#
#=\frac{d}{dx}(e^x)-\frac{d}{dx}(3x)# and
#\frac{d}{dx}(e^x)=e^x#, #\frac{d}{dx}(3x)=3#}
#=e^x-3#............. #eq^n (ii)#
Again,
#\frac{d}{dx}(\cot (x)+x^2)=-\frac{1}{\sin ^2(x)}+2x#
{Applying sum and differnce rule as: #(f\pm g)^'=f^'\pm g^'#
#=\frac{d}{dx}(\cot (x))+\frac{d}{dx}(x^2)# and
and we know the common derivative,
#\frac{d}{dx}(\cot (x))=-\frac{1}{\sin ^2(x)}# and also we know,#\frac{d}
{dx}(x^2)=2x#}
#=-\frac{1}{\sin ^2(x)}+2x# ............. #eq^n (iii)#
Finally,from #eq^n (i), (ii) and (iii)#, we get,
#=(e^x-3)(\cot (x)+x^2)+(-\frac{1}{\sin ^2(x)}+2x)(e^x-3x)#
