Question

differentiation of sin square x + cos square x

Answer 1

Method (1):

= > d/dx(sin^2 x + cos^2 x)

We know that sin^2theta + cos^2theta = 1

= > d/dx(1)

= > 0.



Method(2):

= > d/dx[sin^2 x + cos^2 x]

= > d/dx[sin^2 x] + d/dx[cos^2 x]

= > 2 sinx * d/dx[sinx] + 2 cosx * d/dx[cosx]

= > 2cosxsinx - 2sinxcosx

= > 0.



Hope this helps!

Answer 2

Hey !!!

sin²x + cos²x

dy/dx = d(sinx)²/dx + d(cos²x)/dx

using chain rule

dy/dx = cosx(2sinx ) + (-sinx)(2sinx)

dy/dx = 2sinx×cosx - 2sinx*cosx

dy/dx = 0

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2nd method

As we know

sin²x + cos²x = 1

hence, differentiation of constant term is always zero "0

i.e dy/dx = 0

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Hope it helps you !!!
$.$

@Rajukumar111 ;-)