Knowing that, #g# is the inverse function of #f#, we have,
#g@f"=the Identity fun. "rArr g(f(x))=I(x)=x....(star)#
#rArr d/dx g(f(x))=d/dx(x).#
#"By the Chain Rule, "g'(f(x))f'(x)=1.#
In particular, for #x=3, g'(f(3))f'(3)=1................(ast)#
But, #f(3)=2, and, f'(3)=1/9 :. (ast) rArr g'(2)=9..................(1)#
Now, #G(x)=1/g(x)rArr G'(x)=-(g'(x))/[g(x)]^2..."[the Chain Rule]"#
Hence, for #x=2, G'(2)=-(g'(2))/[g(2)]^2=-9/[g(2)]^2, ........[because, (1)]#
#"Here, to find, "g(2)," we take "x=3" in "(star)" & get, "g(f(3))=3, i.e., g(2)=3," as "f(3)=2.#
So, finally, #G'(2)=-9/3^2=-1.#
Enjoy Maths., and, Spread the Joy!
