# What is derivative of f(x) = 2^(arctg x)?

### 1 Answer | Add Yours

You may notice that the given function is the result of composition of two functions, g(x) = 2^x and h(x) = arctan x, such that:

(goh)(x) = g(h(x)) => g(h(x)) = 2^(h(x)) => g(h(x)) = 2^(arctan x)

Since the function `f(x)` is the result of composition of two functions, you may evaluate its derivative using the chain rule, hence, you need to start differentiation process from outside toward inside, such that:

`f'(x) = (2^(arctan x))'*(arctan x)'`

`f'(x) = (2^(arctan x)*ln 2)*(1/(1 + x^2))`

`f'(x) = (2^(arctan x)*ln 2)/(1 + x^2)`

You may also use, as alternative method, the logarithmic differentiation, such that:

`f(x) = y => y = 2^(arctan x)`

Taking common logarithms both sides, yields:

`ln y = ln (2^(arctan x))`

Using the power property of logarithms yields:

`ln y = arctan x*ln 2`

Differentiation both sides, yields:

`(1/y)*y' = (1/(1 + x^2))*ln 2 `

`y' = y*(ln 2/(1 + x^2))`

Replacing `2^(arctan x)` for y yields:

`y' = (2^(arctan x)*ln 2)/(1 + x^2)`

**Hence, evaluating the derivative of the given function, using either the chain rule, or logarithmic differentiation, yields **`y' = (2^(arctan x)*ln 2)/(1 + x^2).`