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

1 Answer
Nov 28, 2017

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)