Let g:R→R be given by g(x)=e2x+3x+sin⁡x+1 . If g−1 is the inverse function of g ,  then the value of 1g−1'(2) is

Let g−1(x)=f(x)∴f(g(x))=xf′(g(x))⋅g′(x)=1   ⇒f'gx=1g'x    To find f'2, make gx=2⇒ e2x+3x+sinx+1=2 ⇒x=0                       therefore     g0=2now f'2= f'g0= 1g'0 =16   ∵ g'(x)=2e2x+3+cosx ⇒g'0=6 ∴1f′(2)=6