Find derivative using formal definition for `f(x)=1/(3x-1)`

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By definition of derivative 

`f'(x)=lim_(Delta x->0)(f(x+Delta x)-f(x))/(Delta x)`

so for `f(x)=1/(3x-1)` we have

`f'(x)=lim_(Delta x->0)(1/(3x+3Delta x-1)-1/(3x-1))/(Delta x)=lim_(Delta x->0)((3x-1-3x-3Delta x+1)/((3x+3Delta x-1)(3x-1)))/(Delta x)=`

`lim_(Delta x->0)(-3Delta x)/((3x+Delta x-1)(3x-1)Delta x)=lim_(Delta x->0)-3/((3x+3Delta x-1)(3x-1))=`

`-3/(3x-1)^2`

So your solution is `f'(x)=-3/(3x-1)^2`

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By definition of derivative 

`f'(x)=lim_(Delta x->0)(f(x+Delta x)-f(x))/(Delta x)`

so for `f(x)=1/(3x-1)` we have

`f'(x)=lim_(Delta x->0)(1/(3x+3Delta x-1)-1/(3x-1))/(Delta x)=lim_(Delta x->0)((3x-1-3x-3Delta x+1)/((3x+3Delta x-1)(3x-1)))/(Delta x)=`

`lim_(Delta x->0)(-3Delta x)/((3x+Delta x-1)(3x-1)Delta x)=lim_(Delta x->0)-3/((3x+3Delta x-1)(3x-1))=`

`-3/(3x-1)^2`

So your solution is `f'(x)=-3/(3x-1)^2`

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