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,

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

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

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

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

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