Answer to Question #162391 in Calculus for Fiza Aejaz Mandlekar

Question

The derivate of inv function of f:|0,1 |➡️R f(x) =xe^x at x=0.5 is

Answer 1

$f(x)=x*e^x$

$f'(x)=e^x+x*e^x$

$f^{-1}(x) -\text{inverse function }f(x)$

$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$

$f^{-1}(x):x=y*e^y$

$x=y*e^y - \text{this function does not allow you to explicitly express y in x}$

$x\approx{y*(1+y) }$

$y\approx\frac{-1+\sqrt{1+4*x}}{2}$

$f^{-1}(0.5)\approx0.37$

$(f^{-1}(0.5))'=\frac{1}{f'(0.37)}=\frac{1}{e^{0.37}+0.37*e^{0.37}}\approx 0.504$

Answer:0.504