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 $f'(x) = 0$

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