Question

$L$et $f\left(x\right)=\frac{3x+2}{5x-3}$ then ${\left[\frac{d}{dx}\left(f^{-1}\right(x\left)\right)\right]}_{x=f\left(1\right)}$ equals

• A. $\frac{-19}{4}$
• B. $\frac{-4}{19}$
• C. $-9$
• D. $0$

Answer 1

The correct option is B $\frac{-4}{19}$

$f\left(x\right)=\frac{3x+2}{5x-3}$
Let $f^{-1}\left(x\right)=y$
$\therefore x=f\left(y\right)$
$\therefore x=\frac{3y+2}{5y-3}$
$5xy-3x=3y+2$

$5xy-3y=3x+2$

$y[5x-3]=3x+2$

$y=\frac{3x+2}{5x-3}$
$f^{-1}\left(x\right)=\frac{3x+2}{5x-3}$
$\frac{d}{dx}\left[f^{-1}\left(x\right)\right]$

$=\frac{(5x-3)\left(3\right)-(3x+2)\left(5\right)}{{(5x-3)}^2}$ [use quotient rule]

$=\frac{15x-9-15x-10}{{(5x-3)}^2}$

$=\frac{-19}{{(5x-3)}^2}$

$f\left(1\right)=\frac{3\left(1\right)+2}{5\left(1\right)-3}=\frac{5}{2}$

$\therefore {\left[\frac{d}{dx}\left[f^{-1}\left(x\right)\right]\right]}_{x=f\left(1\right)}=\frac{-19}{{(5\times \frac{5}{2}-3)}^2}=\frac{-19}{{\left[\frac{25-6}{2}\right]}^2}=\frac{-19\times 4}{{\left(19\right)}^2}=\frac{-4}{19}$