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f(x)=cos^2x+sin^2x derivative
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-2cos(x)sin(x)+2sin(x)cos(x) = 0
... since f(x) = 1!
did u consider this as basic and composite functions? because they have different steps of solving it
2(cosx)*-sinx + 2(sinx)*cosx -2cosxsinx + 2cosxsinx 0?
it is 0 because your function is the constant function \[f(x)=1\]
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so i dont need to show the working?
don't be hoodwinked in to finding the derivative using chain and product rule, that is a silly waste of time
but i love the chain rule : (
here is the work \[\sin^2(x)+\cos^2(x)=1\] \[\frac{d}{dx}[1]=0\]
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