Question

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

Answer 1

`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)`