# If `f(x) = 1/(4x-1)` determine `f'(x) = lim_(h->0)(f(x+h)-f(x))/h`

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### 1 Answer

The function `f(x) = 1/(4x-1)`

f'(x) can be determined by evaluating `lim_(h->0) (f(x+h)-f(x))/h`

`lim_(h->0) (f(x+h)-f(x))/h`

=> `lim_(h->0) (1/(4*(x+h)-1) - 1/(4x-1))/h`

=> `lim_(h->0) (1/(4*x+4h-1) - 1/(4x-1))/h`

=> `lim_(h->0) (4x - 1 - 4x - 4h + 1)/(h*(4*x+4h-1)*(4x-1))`

=> `lim_(h->0) (-4h)/(h*(4*x+4h-1)*(4x-1))`

=> `lim_(h->0) -4/((4*x+4h-1)*(4x-1))`

substituting h =0

=> `-4/(4x - 1)^2`

**For `f(x) = 1/(4x-1)` , **`f'(x) = -4/(4x - 1)^2`