How do I find the derivative of the function #y=ln (sqrt(x^2-9))#?

1 Answer
Massimiliano
Jan 27, 2015

The derivative of this function is: #y'=x/((x^2-9)#.

It is important to remember which is the derivative of the #y=ln[f(x)]#, which is the derivative of #y=sqrt(f(x))#, which is the derivative of #y=[f(x)]^n# and the theorem of the function of function derivation.

  • The derivative of #y=ln[f(x)]# is #y'=(f'(x))/f(x)#.
  • The derivative of #y=sqrt(f(x))# is #y'=(f'(x))/(2sqrt(f(x)))#.
  • The derivative of #y=[f(x)]^n# is #y'=n[f(x)]^(n-1)f'(x)#.
  • The theorem of the function of function derivation says that the derivative of #y=[f(g(x))]# is #y'=f'(g(x))g'(x)#.

#y'=1/(sqrt(x^2-9))1/(2sqrt(x^2-9))2x=x/(x^2-9)#

There is another way to do this derivative, rememberig one property of the logarithmic function:

#ln(x^n)=nlog(x)#

so:

#y=ln(sqrt(x^2−9))=ln(x^2-9)^(1/2)=(1/2)log(x^2-9)#

and the derivative of this function is easier:

#y'=(1/2)1/(x^2-9)2x=x/(x^2-9)#

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