Question

If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to

• A. $\frac{1}{1+\left(g\right(x)-x{)}^2}$
• B. $\frac{1}{1-\left(g\right(x)-x{)}^2}$
• C. $\frac{1}{2+\left(g\right(x)-x{)}^2}$
• D. $\frac{1}{2-\left(g\right(x)-x{)}^2}$

Answer 1

The correct option is C $\frac{1}{2+\left(g\right(x)-x{)}^2}$

$Lety=f\left(x\right)\Rightarrow x=f^{-1}\left(y\right)$
$thenf\left(x\right)=x+tanx$
$\Rightarrow x=f^{-1}\left(y\right)+tan\left(f^{-1}\right(y\left)\right)$
$\Rightarrow x=g\left(y\right)+tan\left(g\right(y\left)\right)orx=g\left(x\right)+tan\left(g\right(x\left)\right)\cdots \left(i\right)$
Differentiating both sides, then we get
$1=g^{\prime }\left(x\right)+sec2g\left(x\right).g^{\prime }\left(x\right)$
$\therefore g^{\prime }\left(x\right)=\frac{1}{1+se{c}^2\left(g\right(x\left)\right)}=\frac{1}{1+1+ta{n}^2\left(g\right(x\left)\right)}$
$=\frac{1}{2+(x-g(x){)}^2}[fromEq.(i\left)\right]$
$\frac{1}{2+\left(g\right(x)-x{)}^2}$