Question Number 190522 by Rupesh123 last updated on 04/Apr/23
Answered by gatocomcirrose last updated on 05/Apr/23
$$\mathrm{g}'\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{y}\right)\Rightarrow\mathrm{g}''\left(\mathrm{y}\right)=\mathrm{f}'\left(\mathrm{y}\right) \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{cosx}\sqrt{\mathrm{1}+\mathrm{sen}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{g}''\left(\pi/\mathrm{6}\right)=\mathrm{f}'\left(\pi/\mathrm{6}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$