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#