How do you find the derivative of $ln(x^2+1)$ ?

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Answer 1 · 2017-02-12T06:39:40

$frac{d}{dx}(y)=frac{2x}{x^2+1}$

Given equation is $y=ln(x^2+1)$

One fundamental equation of derivative calculus is
$frac{d}{dx}(y)=1/x$ when $y=lnx$

Another important aspect of derivative calculus is the chain rule, that is if $y=f(t)$ and $t=g(x)$ , then
$dy/dx=dy/dt*dt/dx$

Taking $t=x^2+1$ , from chain rule

$dy/dx=\frac{d}{dt}(lnt)*dt/dx$
$\implies dy/dx=1/t*fracddx(x^2+1)$

Substitute for whatever $t$ remains in the equation as $x^2+1$ and you'll get the answer.

How do you find the derivative of ln(x^2+1)ln(x^2+1)?