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))#