How do you differentiate #f(x)=(e^x-3x)(cotx+x^2)# using the product rule?

1 Answer
Mar 1, 2016

Answer:

#\frac{d}{dx}((e^x-3x)(\cot (x)+x^2))=(e^x-3)(\cot (x)+x^2)+(-\frac{1}{\sin ^2(x)}+2x)(e^x-3x)#

Explanation:

#\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)#