Seniors help

I've a doubt and want it to be clearied...

Let us suppose...

(a - 3)² + (b - 4)² + (c - 8)² + (d - 14)² + (e - 19)² = 0

then we've to find a + b + c + d + e...??

Now, clearly, a = 3, b = 4, c = 8, d = 14 and e = 19

and thus, a + b + c + d + e = 48

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Can't we differentiate it on both the sides

i.e., if (a - 3)² + (b - 4)² + (c - 8)² + (d - 14)² + (e - 19)² = 0
so, on differentiating both the sides we get,

2(a - 3) + 2(b - 4) + 2(c - 8) + 2(d - 14) + 2(e - 19) = 0

=> (a - 3) + (b - 4) + (c - 8) + (d - 14) + (e - 19) = 0

or, a + b + c + d + e = 48 ?????

I know u'll take it as a silly doubt.... but can this be done???..... i m in xth so its not silly for me