How do you differentiate #sinx+sinxcosx#?

2 Answers
Jan 31, 2017

Answer:

#cosx + cos^2x - sin^2x#

Explanation:

The derivative of #sinx# is #cosx#.

For #sinxcosx#, we need to use the product rule, so let #a,b# be functions of #x#. Then #(ab)' = a'b + ab'#.

#(sinxcosx)' = (sinx)'cosx + sinx(cosx)' = cosx * cosx + sinx * (-sinx)#

#=cos^2x - sin^2x#.

The final derivative is #cosx + cos^2x - sin^2x#.

Jan 31, 2017

Answer:

#cosx+cos2x#

Explanation:

We require to know the following #color(blue)"standard derivatives"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(sinx)=cosx , d/dx(cosx)=-sinx)color(white)(2/2)|)))#

Also the #color(blue)"trigonometric identity"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(cos2x=cos^2x-sin^2x)color(white)(2/2)|)))#

#sinxcosx" has to be differentiated using the"color(blue)" product rule"#

#"Given "f(x)=g(x).h(x)" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))#

#"here "g(x)=sinxrArrg'(x)=cosx#

#"and "h(x)=cosxrArrh'(x)=-sinx#

#rArrd/dx(sinxcosx)=sinx(-sinx)+cosx(cosx)#

#=cos^2x-sin^2x=cos2x#

#"Thus "d/dx(sinx+sinxcosx)=cosx+cos2x#