How do you differentiate #y=sin^(2)x + cos^(2)x#?

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Answer 1 · 2016-06-04T02:29:19

#dy/dx=0#

From the given
#y=sin^2 x+cos^2 x#

The right side of the equation is #=1#

#y=1#

#dy/dx=d/dx(1)=0#

or we can do it this way.

#y=sin^2 x+cos^2 x#

#dy/dx=2*(sin x)^(2-1)*d/dx(sin x)+2(cos x)^(2-1)d/dx(cos x)#

#dy/dx=2*sin x*cos x+2*cos x*(-sin x)#

#dy/dx=2*sin x*cos x-2*sin x*cos x#

#dy/dx=0#

God bless....I hope the explanation is useful.