How do you differentiate #sqrt(x^2-16)#?

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Answer 1 · 2016-07-26T17:41:59

#d/(dx) sqrt(x^2-16) = x/sqrt(x^2-16)#

Use a combination of the power rule and chain rule to find:

#d/(dx) sqrt(x^2-16)#

#= d/(dx) (x^2-16)^(1/2)#

#= 2x*1/2 (x^2-16)^(-1/2)#

#=x/sqrt(x^2-16)#

The original function has domain #(-oo, -4] uu [4, oo)#

The original function has undefined slope for #x=+-4#, hence the derivative has domain #(-oo, -4) uu (4, oo)#.

Answer 2 · 2016-07-26T17:52:39

#x/sqrt(x^2-16)#.

Let #y=sqrt(x^2-16)={(x-4)(x+4)}^(1/2)#

#:. lny=1/2{ln(x-4)(x+4)}=1/2{ln(x-4)+ln(x+4)}#

#:. d/dx(lny)=1/2{1/(x-4)+1/(x+4)}#

Using the Chain Rule for,

#d/dx(lny)=d/dy(lny)*dy/dx=1/y*dy/dx#

#:. 1/y*dy/dx=1/2{(x+4+x-4))/((x-4)(x+4))#.

#;. dy/dx=(yx)/(x^2-16)=x/(x^2-16)*sqrt(x^2-16)#

#:. y'=x/sqrt(x^2-16)#.