How do you find the derivative of # f(x)=arcsin(4x)+arccos(4x)#?

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Answer 1 · 2017-03-12T17:20:01

#f'(x)=0#

#f(x)=arcsin4x+arccos4x#

Let #u=arcsin4x# and #v=arccos4x#

#4x=sinu# and #4x=cosv#

#(du)/dxcosu=4#

#(du)/dx=4/cosu=4/(sqrt(1-sin^2u))=4/(sqrt(1-16x^2))#

#-(d v)/dxsinv=4#

#(d v)/dx=-4/sinv=-4/(sqrt(1-cos^2v))=-4/(sqrt(1-16x^2))#

#f'(x)=u'+v'==4/(sqrt(1-16x^2))-4/(sqrt(1-16x^2))=0#

Answer 2 · 2017-03-12T17:20:28

#arcsin (x)+arccos(x) = pi/2#.

So #f'(x) = 0#