How do you differentiate #y=ln(10/x)#?
1 Answer
Sep 10, 2016
Explanation:
We need to know that:
#log(A/B)=log(A)-log(B)# #d/dxln(x)=1/x# #d/dx("constant")=0#
Thus:
#y=ln(10/x)#
#y=ln(10)-ln(x)#
#dy/dx=0-1/x#
#dy/dx=-1/x#
