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

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Answer 1 · 2016-03-18T10:50:42

#d/(dx)(e^x-3x)(tanx+x^2)#

= #(e^x-3x)(sec^2x+2x)+(e^x-3)(tanx+x^2)#

According to product rule

#d/(dx)(g(x)xxh(x))=g(x)xxd/(dx)h(x)+d/(dx)g(x)xxh(x)#

As #(dg)/(dx)=d/(dx)(e^x-3x)=e^x-3# and

#(dh)/(dx)=d/(dx)(tanx+x^2)=sec^2x+2x#

Hence #d/(dx)(e^x-3x)(tanx+x^2)#

= #(e^x-3x)(sec^2x+2x)+(e^x-3)(tanx+x^2)#