Question Number 35084 by Raj Singh last updated on 15/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 15/May/18
Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18
$${y}={tan}^{−\mathrm{1}} \left({sinx}+{cosx}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} \:}×\left({cosx}−{sinx}\right) \\ $$$${cosx}>{sinx}\:{in}\left(\mathrm{0},\Pi/\mathrm{4}\right) \\ $$$${so}\:\frac{{dy}}{{dx}}>\mathrm{0}\:\:\:{so}\:{y}\:{is}\:{increasing}\:{function}\:{in}\left(\mathrm{0},\Pi/\mathrm{4}\right) \\ $$