How do you find the first and second derivative of #ln(x^2-4)#?

1 Answer
Oct 2, 2016

Answer:

#f'(x)=(2x)/(x^2-4)#

#f''(x)=(-2(x^2+4))/(x^2-4)^2#

Explanation:

let # f(x)=ln(x^2-4)#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(d/dx(lnx)=1/x)color(white)(a/a)|)))#

and more generally. #color(red)(bar(ul(|color(white)(a/a)color(black)(d/dx(ln(g(x)))=(g'(x))/g(x))color(white)(a/a)|)))#

#rArrf'(x)=(2x)/(x^2-4)#

To find the second derivative, use the #color(blue)"quotient rule"#

Given #f(x)=g(x)/(h(x))" then"#

#color(red)(bar(ul(|color(white)(a/a)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(a/a)|)))......(A)#

here #g(x)=2xrArrg'(x)=2#

and #h(x)=x^2-4rArrh'(x)=2x#

substitute these values into (A)

#f''(x)=((x^2-4).2-2x.2x)/(x^2-4)^2#

#=(2x^2-8-4x^2)/(x^2-4)^2=(-2x^2-8)/(x^2-4)^2=(-2(x^2+4))/(x^2-4)^2#