Prove using the first principal that #d/dx (sqrtcosx) = -sinx/ (2(sqrtcosx)# ?

Question

802 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T06:43:27

Please see below.

As #f(x)=sqrt(cosx)#, #f(x+h)=sqrt(cos(x+h))# and

#(dy)/(dx)=lim_(h->0)(f(x+h)-f(x))/h#

= #lim_(h->0)(sqrt(cos(x+h))-sqrtcosx)/h#

= #lim_(h->0)(sqrt(cos(x+h))-sqrtcosx)/hxx(sqrt(cos(x+h))+sqrtcosx)/(sqrt(cos(x+h))+sqrtcosx)#

= #lim_(h->0)(cos(x+h)-cosx)/(h(sqrt(cos(x+h))+sqrtcosx))#

= #lim_(h->0)(-2sin(x+h/2)sin(h/2))/(h(sqrt(cos(x+h))+sqrtcosx))#

= #lim_(h->0)(sin(h/2)/(h/2))xxlim_(h->0)(-sin(x+h/2))/(sqrt(cos(x+h))+sqrtcosx)#

= #1xx(-sinx)/(2sqrt(cosx))#

= #-sinx/(2sqrt(cosx))#