f: R+∪{0}→R f(x)=arcsin⁡(cos⁡x)+arccos⁡(sin⁡x) then what is the derivative of function f? A)−1x B) 1x c)12x

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f: R+∪{0}→R    f(x)=arcsin⁡(cos⁡x)+arccos⁡(sin⁡x) then what is the derivative of function f?   A)−1x     B) 1x     c)12x

porschomcl · Accepted Answer

Applying the chain rule we get:f′(x)=11−(cos⁡x)2⋅(−sin⁡x)⋅12x−11−(sin⁡x)2⋅(cos⁡x)⋅12x==−12x⋅(sin⁡x|sin⁡x|+cos⁡x|cos⁡x|)So A) is a correct answer, in the interval (0,(π2)2) where both sin⁡x and cos⁡x are positive.

Joseph Fair · Accepted Answer

Shortcut solution:  arcsin⁡(cos⁡(x))+arccos⁡(sin⁡(x))  =arcsin⁡(sin⁡(π2−x))+arccos⁡(cos⁡(π2−x))   Using this transformation, the rest is easy.

Vasquez · Accepted Answer

Hint: f(x)=arcsin⁡(cos⁡x)+π2−arcsin⁡(sin⁡x) =π2+arcsin⁡(cos⁡x)+arcsin⁡(−sin⁡x) Using Proof for the formula of sum of arcsine functions arcsin⁡x+arcsin⁡y, f(x)=π2+arcsin⁡(cos⁡x|sin⁡x|−sin⁡x|cos⁡x|) as here x=cos⁡x,   y=−sin⁡x⇒x2+y2=1 Now use |u|={u  if  u&amp;gt;0−uu≤0 For example if cos⁡x,sin⁡x have the same sign ⇔nπ≤x≤nπ+π2 where n is any integer

f: \mathbb{R}^+ \cup \{0\} \to \mathbb{R} f(x)=\arcsin(\cos \sqrt x)+\arccos(\sin \sqrt

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