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

Answer 1

Ezra Adams

#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.

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Answer 2

Samantha Adkison

To differentiate ( y = \sin^2(x) + \cos^2(x) ), you can use the fact that ( \sin^2(x) + \cos^2(x) = 1 ) for all ( x ). Therefore, the derivative of ( y ) with respect to ( x ) is ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.