Lösung 1.2:2c

Aus Online Mathematik Brückenkurs 2

When we see this expression, we should think "square root of something",

,

and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression =cosx,

ddxcosx=12cosxcosx

where we have used the differentiation rule

ddxx=ddxx12=21x12−1=21x−12=12x.

Thus, we obtain

ddxcosx=12cosx(−sinx)=−sinx2cosx.