Question Number 167221 by rexford last updated on 09/Mar/22
$${The}\:{plane}\:{y}=\mathrm{1}\:{slices}\:{the}\:{surface}\: \\ $$$${z}={arctan}\left(\frac{{x}+{y}}{\mathrm{1}−{xy}}\right) \\ $$$${in}\:{a}\:{curve}\:{C}. \\ $$$${Find}\:{the}\:{slope}\:{of}\:{the}\:{tangent}\:{line}\:{to} \\ $$$${C}\:{at}\:{x}=\mathrm{2} \\ $$
Answered by TheSupreme last updated on 10/Mar/22
$${z}={arctan}\left(\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$${z}\left(\mathrm{2}\right)={arctan}\left(−\mathrm{3}\right) \\ $$$${z}'=−\frac{\mathrm{2}}{\left(\mathrm{1}+\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${z}={z}'\left(\mathrm{2}\right)\:\left({x}−\mathrm{2}\right)+{z}\left(\mathrm{2}\right) \\ $$$${z}\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\left({x}−\mathrm{2}\right)+{arctan}\left(−\mathrm{3}\right) \\ $$
Commented by rexford last updated on 10/Mar/22
$${thank}\:{you} \\ $$