What is the derivative of #sinx+cosx#?

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Answer 1 · 2018-05-29T08:23:25

#cosx-sinx#

#d/dx[sinx+cosx]#

Separate,

#d/dx[sinx]+d/dx[cosx]#

Differentiate,

#cosx-sinx#

Answer 2 · 2018-05-29T10:21:49

#cosx-sinx#

#"using the "color(blue)"standard derivatives"#

#•color(white)(x)d/dx(sinx)=cosx" and "d/dx(cosx)=-sinx#

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