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Question Number 128425 by bramlexs22 last updated on 07/Jan/21

If f(x) = x + tan x and f is inverse of g , then g′(x) is equal to (a) (1/(1+(g(x)−x)^2 )) (b) (1/(1−(g(x)−x)^2 )) (c) (1/(2+(g(x)−x)^2 )) (d) (1/(2−(g(x)−x)^2 ))

$$\:\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\:\mathrm{and}\:\mathrm{f}\:\mathrm{is}\:\mathrm{inverse} \\ $$$$\mathrm{of}\:\mathrm{g}\:,\:\mathrm{then}\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} }\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{c}\right)\:\frac{\mathrm{1}}{\mathrm{2}+\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} }\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{1}}{\mathrm{2}−\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} } \\ $$

Answered by liberty last updated on 07/Jan/21

⇒f(x)=g^(−1) (x); g(x)=f^(−1) (x) ⇒g′(x) = (1/(f ′(f^(−1) (x)))) ; f ′(x)=1+sec^2 x=2+tan^2 x ⇒g ′(x)=(1/(2+(x+tan x−x)^2 )) ⇒g′(x)=(1/(2+(g(x)−x)^2 ))

$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}^{−\mathrm{1}} \left(\mathrm{x}\right);\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{g}'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{f}\:'\left(\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)}\:;\:\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{1}+\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}=\mathrm{2}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\mathrm{g}\:'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}+\left(\mathrm{x}+\mathrm{tan}\:\mathrm{x}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}+\left(\mathrm{g}\left(\mathrm{x}\right)−\mathrm{x}\right)^{\mathrm{2}} } \\ $$

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