How do you differentiate #f(x)=4sqrt((x^2+1)/(x^2-1))#?

1 Answer
Ratnaker Mehta
Jun 17, 2016

#f'(x)=(-8x)/{(x^2-1)^(3/2)(x^2+1)^(1/2)}.#

Explanation:

#y=f(x)=4sqrt{(x^2+1)/(x^2-1)}=4{(x^2+1)/(x^2-1)}^(1/2).#

Taking ln of both sides,
#lny=ln4+1/2{ln(x^2+1)-ln(x^2-1)}#

Diff. both sides w.r.t x,
#d/dxlny=d/dxln4+1/2{d/dxln(x^2+1)-d/dxln(x^2-1)}.#
#d/dylny*dy/dx=0+1/2{1/(x^2+1)*d/dx(x^2+1)-1/(x^2-1)*d/dx(x^2-1)}#
#1/y*dy/dx=1/2[1/(x^2+1)*2x-1/(x^2-1)*2x}=1/2*2x[{x^2-1-x^2-1]/{(x^2-1)(x^2+1)]=-(2x)/{(x^2-1)(x^2+1)}#

Therefore, #dy/dx=f'(x)=y*(-2x)/{(x^2-1)(x^2+1)}=4sqrt{(x^2+1)/(x^2-1)}*(-2x)/{(x^2-1)(x^2+1)}=(-8x)/{(x^2-1)^(3/2)(x^2+1)^(1/2)}.#

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