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# Find the derivative of ln((x^2+1)/(x+3))^0.5

## Answer

$\frac{1}{2}\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}\left(\frac{2x\left(3+x\right)-\left(1+x^2\right)}{\left(3+x\right)\left(1+x^2\right)}\right)$

## Step-by-step explanation

Problem

$\frac{d}{dx}\left(\sqrt{\ln\left(\frac{x^2+1}{x+3}\right)}\right)$
1

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{1}{2}\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}\cdot\frac{d}{dx}\left(\ln\left(\frac{1+x^2}{3+x}\right)\right)$

## Answer

$\frac{1}{2}\ln\left(\frac{1+x^2}{3+x}\right)^{-\frac{1}{2}}\left(\frac{2x\left(3+x\right)-\left(1+x^2\right)}{\left(3+x\right)\left(1+x^2\right)}\right)$

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$\frac{d}{dx}\left(\sqrt{\ln\left(\frac{x^2+1}{x+3}\right)}\right)$

### Main topic:

Differential calculus

~ 0.78 seconds