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

1 Answer
May 29, 2017

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

Explanation:

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

Quotient rule:

#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#