Answer :

Given,

f(x) = x + 1

Therefore fof(x) = f[f(x)]

= f(x) + 1

= (x + 1) + 1

= x + 2

Differentiating with respect to x, we get,







As we know,





= 1x1-1 + 0x0-1

= 1x0 + 0x-1

= 1 + 0

= 1

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