How do I find $f'(3)$ given that $f(x)=sqrt(3x)/(x^2-4)$ ?

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Answer 1 · 2017-05-29T13:56:55

$f'(3)=-31/50$

Let $f(x)=sqrt(3x)/(x^2-4)$

$d/dx((p(x))/(q(x)))=(q(x)p'(x)-p(x)q'(x))/(q(x)^2)$

Let $p(x)=sqrt(3x)$ , then $p'(x)=sqrt3/(2sqrtx)$

Let $q(x)=x^2-4$ , then $q'(x)=2x$

$f'(x)=((sqrt3/(2sqrtx))(x^2-4)-sqrt(3x)(2x))/(x^2-4)^2$

Since we only care about the derivative at $x=3$ , we shall leave simplifying the expression.

$f'(3)=((sqrt3/(2sqrt3))(3^2-4)-sqrt(3(3))(2(3)))/(3^2-4)^2=-31/50$

How do I find f'(3)f'(3) given that f(x)=sqrt(3x)/(x^2-4)f(x)=sqrt(3x)/(x^2-4)?