Question

find the derivative of f(x)= x/sqrt(x^2 - 4)
the answer is -4/(x^2-4)^3/2 but how?

Answer 1

quotient rule plus chain rule

Answer 2

i tried that

Answer 3

i must be doing something wrong

Answer 4

\[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=x,f'(x)=1,g(x)=\sqrt{x^2-4},g'(x)=\frac{x}{\sqrt{x^2-4}}\]

Answer 5

\[\frac{\sqrt{x^2-4}-x\frac{x}{\sqrt{x^2-4}}}{x^2-4}\]
\[\frac{\sqrt{x^2-4}-\frac{x^2}{\sqrt{x^2-4}}}{x^2-4}\]
now last step is to multiply top and bottom by
\[\sqrt{x^2-4}\] and you will have your answer

Answer 6

denominator will be
\[(x^2-4)^{\frac{3}{2}}\]
numerator will be
\[x^2-4-x^2=-4\] and that is what you have above

Answer 7

omg i got it thank you

Answer 8

yw