Haskell types error: Inferred type is less polymorphic than expected

Question

I'm messing around with Haskell at the moment and for the life of me I cannot figure out why the following works.....

square :: (Num a) => a -> a 
square x = x * x 
dx = 0.0000001
deriv1 :: (Fractional a) => (a -> a) -> (a -> a)
deriv1 g = (\x -> ((g (x + 2) - (g x)) / 0.0000001 ))   
main = printf "res==%g %g\n" (square 5.12::Double) ((deriv1 square) 2::Float)

but this doesn't....

square :: (Num a) => a -> a 
square x = x * x 
dx = 0.0000001
deriv1 :: (Fractional a) => (a -> a) -> (a -> a)
deriv1 g = (\x -> ((g (x + 2) - (g x)) / dx ))          
main = printf "res==%g %g\n" (square 5.12::Double) ((deriv1 square) 2::Float)

note I've used dx in the derv1 function this time. I'm new to Haskell so any in depth discussion on types will likely whoosh past me so fast I'll die spinning. It's imperative I have something resembling an imperative answer or it will almost certainly be lost on me this early in my Haskell career.

The error message I'm getting is:

Inferred type is less polymorphic than expected
  Quantified type variable `a' is mentioned in the environment:
    dx :: a (bound at sicp-1.40.hs:12:0)
When trying to generalise the type inferred for `deriv1'
  Signature type:     forall a. (Fractional a) => (a -> a) -> a -> a
  Type to generalise: (a -> a) -> a -> a
In the type signature for `deriv1'
When generalising the type(s) for `deriv1'

Answer 1

You get the error because of the monomorphism restriction. Since you don't give a type signature to dx it ends up inferred as Double in this case. You can either give an explicit, polymorphic signature like

dx :: Fractional a => a
dx = 0.0000001

or you can disable the monomorphism restriction by including this line at the top of your source file

{-# LANGUAGE NoMonomorphismRestriction #-}