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

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Answer 1 · 2016-10-23T16:42:14

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

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

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