How do you differentiate #sinx+sinxcosx#?

Question

Christopher Allers · Accepted Answer

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

Emily Ammerman · Answer

#cosx+cos2x# 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#

Isabella Ackerman · Answer

undefined To differentiate $ \sin(x) + \sin(x)\cos(x) $, you can use the sum rule and the product rule of differentiation.

First, differentiate $ \sin(x) $ with respect to $ x $, which gives $ \cos(x) $.

Next, for the term $ \sin(x)\cos(x) $, use the product rule: differentiate $ \sin(x) $ to get $ \cos(x) $ and then differentiate $ \cos(x) $ to get $ -\sin(x) $, and multiply these results together.

Finally, combine the derivatives of both terms to get the overall derivative of the expression.