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Mathematics 22 Online
OpenStudy (anonymous):

f(x)=cos^2x+sin^2x derivative

OpenStudy (rulnick):

-2cos(x)sin(x)+2sin(x)cos(x) = 0

OpenStudy (rulnick):

... since f(x) = 1!

OpenStudy (anonymous):

did u consider this as basic and composite functions? because they have different steps of solving it

OpenStudy (anonymous):

2(cosx)*-sinx + 2(sinx)*cosx -2cosxsinx + 2cosxsinx 0?

OpenStudy (anonymous):

it is 0 because your function is the constant function \[f(x)=1\]

OpenStudy (anonymous):

so i dont need to show the working?

OpenStudy (anonymous):

don't be hoodwinked in to finding the derivative using chain and product rule, that is a silly waste of time

OpenStudy (anonymous):

but i love the chain rule : (

OpenStudy (anonymous):

here is the work \[\sin^2(x)+\cos^2(x)=1\] \[\frac{d}{dx}[1]=0\]

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