Question

How do you differentiate `y= (3+2^x)^x`?

Answer 1

`dy/dx = (3+2^x)^x((ln(2)x2^x)/(3+2^x) + ln (3+2^x))`

Explanation:

The simplest method is to use what is known as logarithmic differentiation. We take (Natural) logarithms of both sides and the use implicit differentiation:

`y = (3+2^x)^x`

Taking logarithms we get:

`lny = ln {(3+2^x)^x}`
`lny = xln (3+2^x)`

Then we can differentiate the LHS Implicitly, and the RHS using the product rule to get:
`1/ydy/dx = (x)(d/dxln (3+2^x)) + (ln (3+2^x))(1)`
`:. 1/(3+2^x)^x dy/dx = xd/dxln (3+2^x) + ln (3+2^x) ... [1]`

To deal with `d/dxln (3+2^x)` we use the chain rule:

`\ \ \ \ \ d/dxln (3+2^x) = 1/(3+2^x) d/dx(3+2^x)`
`:. d/dxln (3+2^x) = 1/(3+2^x) ln(2)2^x`
`:. d/dxln (3+2^x) = (ln(2)2^x)/(3+2^x)`

Substituting the last result into[1] we get:

`1/(3+2^x)^x dy/dx = (ln(2)x2^x)/(3+2^x) + ln (3+2^x)`
`dy/dx = (3+2^x)^x((ln(2)x2^x)/(3+2^x) + ln (3+2^x))`