どのようにsqrt(cos(x ^ 2 + 2))+ sqrt(cos ^ 2x + 2)を微分しますか?

どのようにsqrt(cos(x ^ 2 + 2))+ sqrt(cos ^ 2x + 2)を微分しますか?
Anonim

回答:

#(dy)/(dx)=(xsen(x ^ 2 + 2)+ sen(x + 2))/(sqrtcos(x ^ 2 + 2)+ sqrt(cos ^ 2(x + 2)))#

説明:

#(dy)/(dx)= 1 /(2sqrtcos(x ^ 2 + 2)+ sqrt(cos ^ 2(x + 2)))* sen(x ^ 2 + 2)* 2x + 2sen(x + 2) )#

#(dy)/(dx)=(2xsen(x ^ 2 + 2)+ 2sen(x + 2))/(2sqrtcos(x ^ 2 + 2)+ sqrt(cos ^ 2(x + 2)))#

#(dy)/(dx)=(cancel2(xsen(x ^ 2 + 2)+ sen(x + 2)))/(cancel2sqrtcos(x ^ 2 + 2)+ sqrt(cos ^ 2(x + 2)) ))#

#(dy)/(dx)=(xsen(x ^ 2 + 2)+ sen(x + 2))/(sqrtcos(x ^ 2 + 2)+ sqrt(cos ^ 2(x + 2)))#