π§(π₯, π¦) = π(π₯, 1) = π₯π ππ(π₯ − 1) 2 + π₯2 =β« πΉ ′ π₯ (1,1,0) = π(1, π¦) = π₯π¦π ππ(π₯ − π¦) , (1,1,0) 1 + π₯ 2 +π¦ 2 ππ§ ((sinβ‘(π₯ − 1) + π₯πππ (π₯ − 1)(2 + π₯ 2 ) − 2π₯ 2 (sinβ‘(π₯ − 1) 1 = = (2 + π₯ 2 )2 ππ₯ 3 π¦π ππ(1 − π¦) 2 + π¦2 =β« πΉ ′ π¦ (1,1,0) = πΉ ′ π§ (1,1,0) = ππ§ ((sinβ‘(1 − π¦) − π¦πππ (1 − π¦)(2 + π¦ 2 ) − 2π¦ 2 (sinβ‘(1 − π¦) 1 = =− 2 2 (2 + π¦ ) ππ¦ 3 ππ§ =1 ππ₯ We have the tangent plane: πΉ ′ π§ (π₯, π¦, π§)(π§ − π§0 ) = πΉ ′ π₯ (π₯, π¦, π§)(π₯ − π₯0 ) + πΉ ′ π¦ (π₯, π¦, π§)(π¦ − π¦0 ) =β« 1(π§ − 0) = 1 1 (π₯ − 1) − (π¦ − π¦0 ) 3 3 1 1 =β« π§ = π₯ − π¦ 3 3