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

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Answer 1 · 2017-03-08T14:04:29

#(df)/(dx)=(e^x+1)(cot^2x+x)+(1-2cotxcsc^2x)(e^x+x)#

Product rule states if #f(x)=g(x)h(x)#

then #(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)#

Here we have #g(x)=e^x+x# and #h(x)=cot^2x+x#

and as #(dg)/(dx)=e^x+1#

and #(dh)/(dx)=2cotx xx -csc^2x+1=-2cotxcsc^2x+1#

hence

#(df)/(dx)=(e^x+1)(cot^2x+x)+(1-2cotxcsc^2x)(e^x+x)#