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

##### 3 Answers

#### Explanation:

Use the quotient rule

Let

Let

Simplify using exponent rules

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

#### Explanation:

Let

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 #

#### Explanation:

The derivative of

#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*