How do you find the derivative of #sinx(sinx+cosx)#?

1 Answer
Jim G.
Apr 10, 2018

#sin2x+cos2x#

Explanation:

#"differentiate using the "color(blue)"product rule"#

#"Given "y=g(x)h(x)" then"#

#dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"#

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

#h(x)=sinx+cosxrArrh'(x)=cosx-sinx#

#rArrdy/dx=sinx(cosx-sinx)+cosx(sinx+cosx)#

#color(white)(rArrdy/dx)=sinxcosx-sin^2x+sinxcosx+cos^2x#

#color(white)(rArrdy/dx)=2sinxcosx+cos^2x-sin^2x#

#color(white)(rArrdy/dx)=sin2x+cos2x#

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