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What is the derivative of #y=2^-x#?

Answer 1

William Abdalla

# dy/dx = -ln2*2^(-x) #

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) #

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Answer 2

Samantha Adkison

The derivative of y = 2^(-x) is dy/dx = -ln(2) * 2^(-x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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