If  f(x)=x+1, find  \frac{\mathrm{d} }{\mathrm{d} x}( f\! o\! f)(x).

 

 

 

 
 
 
 
 

Answers (1)

Given f(x)=x+1\, \, \, \, -(1)

f(f(x))  is same as  (f\!o\!f)(x)

i.e  f(f(x))=f(x)+1    [applying f(f(x)) in equation (1)]

                        =x+1+1

                        =x+2

\therefore f(f(x)) = x+2  which is same as f\!o\!f(x)\, \, -(2)

Differentiating equation (2) on both LHS and RHS:

\frac{\mathrm{d} }{\mathrm{d} x}( f\! o\! f)(x)=\frac{\mathrm{d} }{\mathrm{d} x}(x+2)

                        =\frac{\mathrm{d} }{\mathrm{d} x}(x)+\frac{\mathrm{d} }{\mathrm{d} x}(2)

                        =1+0

                        =1

 

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