The-plane-y-1-slices-the-surface-z-arctan-x-y-1-xy-in-a-curve-C-Find-the-slope-of-the-tangent-line-to-C-at-x-2-

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(((x+1)/(1−x))) z(2)=arctan(−3) z′=−(2/((1+(((1+x)/(1−x)))^2 )(1−x)^2 )) z=z′(2) (x−2)+z(2) z = −(1/5)(x−2)+arctan(−3)

$${z}={arctan}\left(\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$${z}\left(\mathrm{2}\right)={arctan}\left(−\mathrm{3}\right) \\ $$$${z}'=−\frac{\mathrm{2}}{\left(\mathrm{1}+\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${z}={z}'\left(\mathrm{2}\right)\:\left({x}−\mathrm{2}\right)+{z}\left(\mathrm{2}\right) \\ $$$${z}\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\left({x}−\mathrm{2}\right)+{arctan}\left(−\mathrm{3}\right) \\ $$

Commented by rexford last updated on 10/Mar/22

$${thank}\:{you} \\ $$

Facebook Tweet Pin

The-plane-y-1-slices-the-surface-z-arctan-x-y-1-xy-in-a-curve-C-Find-the-slope-of-the-tangent-line-to-C-at-x-2-

Leave a Reply Cancel reply