Question

( begin{aligned} f(x) &=x^{x} g(x) &=f^{-1}(x) Rightarrow f(g(x))=x-0 & Rightarrow f^{prime}(g(x)) g^{prime}(x)=1 end{aligned} )
( begin{aligned} & Rightarrow f^{prime}(g(x))^{g(x)} & Rightarrow g^{prime}(x)=frac{1}{F^{prime}(g(x))} end{aligned} )
( f^{2}(x)=x^{x} )
( begin{aligned} Rightarrow & log f(x)=x log ^{y} Rightarrow & frac{1}{f(x)} f^{prime}(x)=frac{x}{x}+log x & f^{prime}(x)=f(x)[f+log x) F^{prime}(g(x))=& F(g(x))[1+log (g(x)) & f^{prime}(g(x))=x[1+log (g(x)] therefore &left[g^{prime}(x)=frac{1}{x(1+log (g(x))}right) end{aligned} )

# they for the (o. a) and le Let f(x) = x*, x € (0,00) and let g(x) be inverse of f(x), then g'(x) must be (A) *(1 + logx) (B) x(1 + logg(x)) 1 (C) (D) non-existent x(1+ logg(x))

Solution