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
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)