Question

What is the derivative of? : `d/dx (x^2 + 1)^(x)`

Answer 1

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

Explanation:

Let:

`y = (x^2 + 1)^(x)`

Taking natural logarithms of both sides:

`lny = ln {(x^2 + 1)^(x)}`

And using the properties of logarithms we have:

`lny = xln (x^2 + 1)`

We now differentiate wrt `x` and apply the product rule:

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

Then using the chain rule this becomes:

`1/y \ dy/dx = x * 1/(x^2+1) * 2x + 1 * ln(x^2+1)`

`:. dy/dx = y{(2x^2)/(x^2+1) + ln(x^2+1) }`

`:. dy/dx = ((2x^2)/(x^2+1) + ln(x^2+1) )(x^2 + 1)^(x)`