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Q.
If $f(x)=(2 x-3 \pi)^5+\frac{4}{3} x+\cos x$ and $g$ is the inverse function of $f$, then $g^{\prime}(2 \pi)$ is equal to
Continuity and Differentiability
Solution:
$ y=(2 x-3 \pi)^5+\frac{4}{3} x+\cos x$
$\frac{d y}{d x}=5(2 x-3 \pi)^4+\frac{4}{3}-\sin x $
$\therefore g^{\prime}(y)=\frac{d x}{d y}=\frac{1}{5(2 x-3 \pi)^4+\frac{4}{3} x-\sin x}$
when $y=2 \pi$
$2 \pi=(2 x-3 \pi)^5+\frac{4}{3} x+\cos x$
$x=\frac{3 \pi}{2}$ is the only solution
$\therefore g ^{\prime}(2 \pi)=\frac{1}{1+\frac{4}{3}}=\frac{3}{7}$
Alternatively: $( g \circ f )( x )= x \Rightarrow g ^{\prime}\left( f ( x ) \cdot f ^{\prime}( x )=1\right.$
$g^{\prime}\left(f(x)=\frac{1}{f^{\prime}(x)}\right.$
Putting $f(x)=2 \pi$
$ 2 \pi=(2 x-3 \pi)^5+\frac{4}{3} x+\cos x $
$\Rightarrow x=\frac{3 \pi}{2} \text { (It is only } x=\frac{3 \pi}{2} \text { as f is one-one function) } $
$\therefore f \left(\frac{3 \pi}{2}\right)=2 \pi $
$\Rightarrow g ^{\prime}(2 \pi)=\frac{1}{ f ^{\prime}\left(\frac{3 \pi}{2}\right)}=\frac{1}{7 / 3}=\frac{3}{7} \left(\because f ^{\prime}( x )=5(2 x -3 \pi)^4+4 / 3-\sin x \right)$