Question

( EN ): Seja f( x ) = x + ln( x ) ; x > 0. g é a inversa de f. Calcule g" ( 1 ) = ?

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Propriedade~da~func_{\!\!,}\tilde ao~inversa}\\\sf sejam~f(x)~e~g(x)~duas~func_{\!\!,}\tilde oes\\\sf bijetivas~quaisquer.\\\sf Se~ g(x)~\acute e~a~inversa~de~f(x)\\\sf ent\tilde ao~f[g(x)]=x~e~g[f(x)]=x\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Derivada~da ~func_{\!\!,}\tilde ao~composta}\\\sf f[g(x)]'= f'[g(x)]\cdot g'(x).\\\sf f(x)=x+\ell n(x)\\\sf f'(x)=1+\dfrac{1}{x}\\\sf f''(x)=-\dfrac{1}{x^2}\\\sf Se~g(x)~\acute e~inversa~de~f(x)\\\sf ent\tilde ao~g[f(x)]=x,ou~seja,g[f(x)]'=g'[f(x)]\cdot f'(x)=1\\\rm da\acute i\\\sf g[f(x)]''=g"[f(x)]\cdot f'(x)\cdot f'(x)+g'(x)\cdot f''(x)=0\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm C\acute alculos~auxiliares:}\\\sf f(1)=1+\ell n(1)=1+0=1\\\sf f'(1)=1+\dfrac{1}{1}=1+1=2\\\sf f''(1)=-\dfrac{1}{1^2}=-\dfrac{1}{1}=-1\\\sf g'[f(1)]\cdot f'(1)=1\\\sf g'(1)\cdot2=1\\\sf g'(1)=\dfrac{1}{2}\end{array}}$

$\boxed{\begin{array}{l}\sf g''[f(1)]\cdot f'(1)\cdot f'(1)+g'(1)\cdot f''(1)=0\\\sf g''(1)\cdot 2\cdot 2+\dfrac{1}{2}(-1)=0\\\sf 4g''(1)-\dfrac{1}{2}=0\\\sf 8g''(1)-1=0\\\sf 8g''(1)=1\\\sf g''(1)=\dfrac{1}{8}\end{array}}$