Question

What is the derivative of `e^x/2^x`?

Answer 1

`(e/2)^x(1-ln 2)`

Explanation:

Use the quotient rule `(u/v)' = (v*u' - u*v')/v^2`
Let `u = e^x`, `u' = e^x`
Let `v = 2^x`, `v' = 2^x ln 2`

`(u/v)' = (2^x*e^x - e^x*2^x ln 2)/(2^x)^2 = ((2^xe^x)(1-ln 2))/(2^(2x))`

Simplify using exponent rules `x^m/x^(2m) = 1/x^(2m-m) = 1/x^m`:

`((2^xe^x)(1-ln 2))/(2^(2x)) = (e^x(1-ln 2))/ 2^x = (e/2)^x(1-ln 2)`

Answer 2

`d/dx e^x/2^x = (1 - ln2)e^x/2^x`

Explanation:

Let `y = e^x/2^x`

Then take (Natural) logarithms of both sides; and use the rules of logs:

`:. ln y = ln {e^x/2^x}`
`" " = ln (e^x) - ln(2^x)`
`" " = xln (e) - xln(2)`
`" " = x - xln(2)`
`" " = x(1 - ln(2))`

Now, differentiate implicitly:

`1/ydy/dx \ = (1 - ln2)`
`:. dy/dx = (1 - ln2)y`
`" " = (1 - ln2)e^x/2^x`

Answer 3

`(e/2)^x(1-ln2)`

Explanation:

`e^x/2^x = (e/2)^x`

The derivative of `a^x` with respect to `x` for any number `a` is:

`d/dxa^x=a^xln(a)`

Therefore:

`d/dx(e/2)^x = (e/2)^xln(e/2)`

`color(white)"XXXXX-" = (e/2)^x(lne-ln2)`
`color(white)"XXXXX-" = (e/2)^x(1-ln2)`

Final Answer