Question Number 92876 by M±th+et+s last updated on 09/May/20
$$\frac{{d}}{{dx}}\left(\sqrt{\mathrm{1}−{x}}+\sqrt{{x}−\mathrm{2}}\right) \\ $$
Commented by i jagooll last updated on 09/May/20
$$=\:\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{x}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\sqrt{\mathrm{x}−\mathrm{2}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}}}\right\}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\sqrt{\mathrm{1}−\mathrm{x}}−\sqrt{\mathrm{x}−\mathrm{2}}}{\sqrt{\mathrm{3x}−\mathrm{x}^{\mathrm{2}} −\mathrm{2}}}\right\} \\ $$
Commented by M±th+et+s last updated on 09/May/20
$${in}\:{this}\:{equation}\:{x}\notin\mathbb{R} \\ $$$${so}\:{is}\:{it}\:{right}\:{if}\:{we}\:{derivative}\:{it}\:{with} \\ $$$${this}\:{way}?? \\ $$
Commented by i jagooll last updated on 09/May/20
yes. correct