Question Number 115260 by bobhans last updated on 24/Sep/20
$${Let}\:{f}\:{be}\:{a}\:{real}\:{valued}\:{function}\:{defined} \\ $$$${on}\:{the}\:{interval}\:\left(−\mathrm{1},\mathrm{1}\right)\:{such}\:{that}\: \\ $$$${e}^{−{x}} .{f}\left({x}\right)=\mathrm{2}+\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\sqrt{{t}^{\mathrm{4}} +\mathrm{1}}\:{dt}\:\forall{x}\in\left(−\mathrm{1},\mathrm{1}\right) \\ $$$${and}\:{let}\:{g}\:{be}\:{the}\:{inverse}\:{function}\:{of}\:{f} \\ $$$$.\:{Find}\:{the}\:{value}\:{of}\:{g}'\left(\mathrm{2}\right). \\ $$
Commented by PRITHWISH SEN 2 last updated on 24/Sep/20
Answered by john santu last updated on 24/Sep/20
$${Differentiating}\:{given}\:{equation}\:{we} \\ $$$${get}\:{e}^{−{x}} .{f}\:'\left({x}\right)−{e}^{−{x}} .{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$${since}\:\left({g}\circ{f}\right)\left({x}\right)={x}\:{as}\:{g}\:{is}\:{inverse}\:{of}\:{f} \\ $$$$\Rightarrow\:{g}\left({f}\left({x}\right)\right)={x}\Rightarrow{f}\:'\left({x}\right).{g}\:'\left({f}\left({x}\right)\right)=\mathrm{1} \\ $$$$\Rightarrow{g}\:'\left({f}\left(\mathrm{0}\right)\right)=\frac{\mathrm{1}}{{f}\:'\left(\mathrm{0}\right)}\Rightarrow{g}\:'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{{f}\:'\left(\mathrm{0}\right)} \\ $$$$\left(\:{here}\:{f}\left(\mathrm{0}\right)=\mathrm{2}\:{obtained}\:{from}\right. \\ $$$$\left.{given}\:{equation}.\right) \\ $$$${put}\:{x}=\mathrm{0}\:{we}\:{get}\:{f}\:'\left(\mathrm{0}\right)=\mathrm{3}. \\ $$$$\therefore\:{g}\:'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by mindispower last updated on 25/Sep/20
$$\left.{g}'\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{{f}'\left({x}\right)},{f}'\left({x}\right)={e}^{{x}} \left(\mathrm{2}+\int_{\mathrm{0}} ^{{x}} \sqrt{\mathrm{1}+{t}^{\mathrm{4}} }\right)+\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}{e}^{{x}} \right) \\ $$$${g}'\left({f}\left(\mathrm{0}\right)\right)=\frac{\mathrm{1}}{{f}'\left(\mathrm{0}\right)}={g}'\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$