What is the derivative of #y=2^-x#?
1 Answer
Mar 20, 2017
# dy/dx = -ln2*2^(-x) #
Explanation:
We have
# y = 2^(-x) #
Take Natural logarithms of both sides and use rules of logs:
# ln y = ln2^(-x) #
# \ \ \ \ \ \ = (-x)ln2 #
# \ \ \ \ \ \ = -xln2 #
Differentiate Implicitly:
# 1/ydy/dx = -ln2 #
# :. dy/dx = (y)(-ln2) #
# " " = -ln2*2^(-x) #
