Question

What is the derivative of `ln(x^2+1)^(1/2)`?

Answer 1

`d/dx{ ln(x^2+1)^(1/2) } = (x)/(x^2+1)`

Explanation:

I am assuming by `ln(x^2+1)^(1/2)` that you mean
`ln(sqrt(x^2+1))` rather than `sqrtln(x^2+1)`

By the chain rule, `d/dxf(g(x)) =f'(g(x))g'(x)` or,`dy/dx=dy/(du)(du)/dx`

So `d/dx{ ln(x^2+1)^(1/2) } = d/dx{ 1/2ln(x^2+1) }` (by the power log rule)

`:. d/dx{ ln(x^2+1)^(1/2) } = 1/2d/dx{ ln(x^2+1) }`
`:. d/dx{ ln(x^2+1)^(1/2) } = 1/2{ 1/(x^2+1)d/dx(x^2+1)}` (by the chain rule)
`:. d/dx{ ln(x^2+1)^(1/2) } = 1/2{ 1/(x^2+1)(2x)}`
`:. d/dx{ ln(x^2+1)^(1/2) } = { 1/(x^2+1)(x)}`

Hence,
`:. d/dx{ ln(x^2+1)^(1/2) } = (x)/(x^2+1)`