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2 is the answer.

PS: Click the image for better view :). (#NEW UPDATE)

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To find the essential prime implicants of F(A,B,C,D)=∑m(0,1,5,7,10,14,15), we can use a Karnaugh map:

CD\AB 00 01 11 10
     -------------
  00|  1  1  1  0
  01|  0  1  0  0
  11|  1  1  0  1
  10|  0  0  0  0

From the K-map, we can see that there are four prime implicants:

  • A'BC'D'
  • AB'C'D'
  • AB'CD
  • ABCD

To determine the essential prime implicants, we need to check if any minterms can only be covered by a single prime implicant. We can use the Petrick's method for this:

F(A,B,C,D) = A'BC'D' + AB'C'D' + AB'CD + ABCD
            = (A'BC'D' + AB'C'D') + (AB'CD + ABCD)
            = (A'BC' + AB'C')D' + (AB'+AB)CD

The first term has the product of sums form (A'BC' + AB'C'), which can be written as (A'+B')C'. The second term has the sum of products form (AB'+AB)CD, which can be written as AB'C+ABC'.

Therefore, the simplified expression is:

F(A,B,C,D) = (A'+B')C'D' + AB'C+ABC'

From this expression, we can see that the essential prime implicants are A'BC'D' and AB'C'D'. The other two prime implicants are not essential because they can be covered by a combination of the essential prime implicants. Therefore, the number of essential prime implicants is 2.

 

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total prime implicates are 6:

a'b'c' + a'c'd + a'bd +bcd + abc + acd'

two of them are essential prime implicates

a'b'c' and acd'

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Hi, The answer is as said by others is 2 Essential Prime Implicants!
Why you ask? 
Well, going by the definition - "Essential prime implicant is an prime implicant that cover an output of the function that no combination of other prime implicants is able to cover."

And if you still couldn't get it! the catch is... notice the K-Map in the picture posted by Hemant Sir, Do you Notice something peculiar at m0 and m10 subcells? 
Well, allow me to simplify it again, notice that in other pairs each one can form LIKE 2 MORE PAIRS, just like a cheating partner!
but those 1's at m0 and m10 position are loyal to their partners! 
Hence out of all the couples in K-Map, we find 2 such Boolean expression where at least 1 person is till uncoupled, hence these expressions are called Essential Prime Implicants
Therefore 2

​​​​​​​Sorry for being so casual in answer, I teach little kids! i am kinda use to explaining things with real world examples

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