Question

How do you differentiate `y=lnsqrt(1+x^2)`?

Answer 1

the derivative is `dy/dx=x/(1+x^2)`

Explanation:

Here we use the chain rule twice

`y=ln(u(x))`

So `dy/dx=(u'(x))/(u(x))`
here `u(x)=sqrt(1+x^2)`
so `u(x)=1/(2sqrt(1+x^2)) *2x`

So `dy/dx = (1/sqrt(1+x^2)) * 1/(2sqrt(1+x^2)) *2x=x/(1+x^2)`