How do you find the first and second derivative of $h(x)=sqrt(x^2+1)$ ?

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Answer 1 · 2017-03-14T05:43:28

$h''(x)=1/((x^2+1)sqrt(x^2+1))$

Since:

$f(x)=sqrt(g(x))->f'(x)=1/(2sqrt(g(x)))*g'(x)$

the first derivative is:

$h'(x)=1/(cancel2sqrt(x^2+1))*cancel2x$

$=x/(sqrt(x^2+1))$

Since:

$h(x)=g(x)/(f(x))->h'(x)=(g'(x)*f(x)-g(x)*f'(x))/(f^2(x))$

the second derivative is:

$h''(x)=(1*sqrt(x^2+1)-x*1/(cancel2sqrt(x^2+1))*cancel2x)/(sqrt(x^2+1))^2$

$=(sqrt(x^2+1)-x^2/sqrt(x^2+1))/(x^2+1)$

$=(cancelx^2+1-cancelx^2)/((x^2+1)sqrt(x^2+1))$

$=1/((x^2+1)sqrt(x^2+1))$

How do you find the first and second derivative of h(x)=sqrt(x^2+1)h(x)=sqrt(x^2+1)?