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,

= 1x^{1-1} + 0x^{0-1}

^{0}+ 0x

^{-1}

= 1 + 0

= 1

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