How do you find the derivative of #y=5/(2x)^2+2cosx#?
1 Answer
Nov 24, 2016
Explanation:
Once we rewrite, we will need the rules:
#d/dxcf(x)=cd/dxf(x)# #d/dxx^n=nx^(n-1)# #d/dxcosx=-sinx#
So first we see that
#y=5/(4x^2)+2cosx#
#y=5/4x^-2+2cosx#
So then
#dy/dx=5/4(-2)x^-3+2(-sinx)#
#dy/dx=5/(2x^3)-2sinx#
