Mathematics - Differentiation Question with Solution | TestHub

If $f(x)=e$, then $x{e}^x=e⟹x=1$, so $g(e)=1$.

We find derivatives of $f(x)$ at $x=1$: $f^{\prime }(x)=e^x(1+x)$,

$f^{\prime \prime }(x)=e^x(2+x)$, $f^{\prime \prime \prime }(x)=e^x(3+x)$.$f^{\prime }(1)=2e$,

$f^{\prime \prime }(1)=3e$, $f^{\prime \prime \prime }(1)=4e$.

Using inverse function derivative formulas: $g^{\prime }(y)=\frac{1}{f^{\prime }(x)}$, $g^{\prime \prime }(y)=-\frac{f^{\prime \prime }(x)}{(f^{\prime }(x){)}^3}$, $g^{\prime \prime \prime }(y)=-\frac{f^{\prime \prime \prime }(x)f^{\prime }(x)-3(f^{\prime \prime }(x){)}^2}{(f^{\prime }(x){)}^5}$.

P. $g^{\prime }(e)=\frac{1}{f^{\prime }(1)}=\frac{1}{2e}$.

Q.$g^{\prime \prime }(e)=-\frac{f^{\prime \prime }(1)}{(f^{\prime }(1){)}^3}=-\frac{3e}{(2e{)}^3}=-\frac{3}{8e^2}$.

R.$g^{\prime \prime \prime }(e)=-\frac{f^{\prime \prime \prime }(1)f^{\prime }(1)-3(f^{\prime \prime }(1){)}^2}{(f^{\prime }(1){)}^5}=-\frac{(4e)(2e)-3(3e{)}^2}{(2e{)}^5}=-\frac{8e^2-27e^2}{32e^5}=\frac{19}{32e^3}$.

S.$\frac{g^{\prime }(e)}{g^{\prime \prime }(e)}=\frac{1/(2e)}{-3/(8e^2)}=-\frac{4e}{3}$.

Thus, P-(1), Q-(2), R-(3), S-(4).