How do you differentiate #f(x)=4/sqrt(x-2)# using the quotient rule?

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Answer 1 · 2016-01-01T17:19:41

#f'(x)=(-2)/(x-2)^(3/2)#

#d/dx[f(x)/(g(x))]=(g(x)*f'(x)-f(x)*g'(x))/([g(x)]^2#.

So in this case:

#d/dx[4]=0#

#d/dx[sqrt(x-2)]=1/2(x-2)^(-1/2)*(1)#

#f'(x)=(sqrt(x-2) * (0)-4(1/2)(x-2)^(-1/2) * (1))/(x-2)#

#=(-2(x-2)^(-1/2))/(x-2)#

#=(-2)/(x-2)^(3/2)#