I solved all the questions with explanations. Please tell me if something is unclear.All solutions are in one file. Docx and pdf files are identical.

We need partial derivatives to find the normal vector:
𝜕𝑧
𝜕𝑧
= 𝑒 𝑥 = 𝑒 0 = 1,
= 1 + 2𝑦 = 11,
𝜕𝑥
𝜕𝑦
so the equation is 1(𝑥 − 0) + 11(𝑦 − 5) − 1(𝑧 − 40) = 0, which gives
𝑧 = 𝒙 + 𝟏𝟏𝒚 − 𝟏𝟓.

Again
𝜕𝑧
𝜕𝑧
= −2𝑥 + 3 = −7,
= 4𝑦 + 2 = 22,
𝜕𝑥
𝜕𝑦
so the equation is −7(𝑥 − 5) + 22(𝑦 − 5) − 1(𝑧 − 51) = 0, which gives
𝑧 = −𝟕𝒙 + 𝟐𝟐𝒚 − 𝟐𝟒.

The gradient consists of three partial derivatives, i.e.
∇𝑓(𝑥, 𝑦, 𝑧) = 〈𝒆𝟒𝒚 𝐬𝐢𝐧(𝟑𝒛) , 𝟒𝒙𝒆𝟒𝒚 𝐬𝐢𝐧(𝟑𝒛) , 𝟑𝒙𝒆𝟒𝒚 𝐜𝐨𝐬(𝟑𝒛)〉.

(𝒂) ∇𝑓(𝑥, 𝑦) = 〈

(𝒃) ∇𝑓(0.6,9) = 〈
(𝒄) 𝑓𝑢 (𝑥, 𝑦) = 0 ∙
so 𝑓𝑢 (0.6,9) =

𝟏

𝐬𝐞𝐜 𝟐 𝒙
𝟐√𝐭𝐚𝐧 𝒙 + 𝒚
𝐬𝐞𝐜 𝟐 𝟎. 𝟔

𝟐√𝐭𝐚𝐧 𝟎. 𝟔 + 𝟗

sec 2 𝑥
2√tan 𝑥 + 𝑦

+2∙

𝟏

,

𝟐√𝐭𝐚𝐧 𝒙 + 𝒚
,

〉.

𝟏
𝟐√𝐭𝐚𝐧 𝟎. 𝟔 + 𝟗
1

2√tan 𝑥 + 𝑦

=

〉.

1
√tan 𝑥 + 𝑦

,

.

√𝐭𝐚𝐧 𝟎.𝟔+𝟗

[I suppose we should give exact answers, not decimal approximations]

The gradient at the given point is a normal vector, and it is equal to 〈−30, − 512, 64〉,
So, the equation is −30(𝑥 + 10) − 512(𝑦 + 8) + 64(𝑧 − 2) = 0, which simplifies to
𝟏𝟓
𝟏𝟏𝟑𝟏
𝑧=
𝒙 + 𝟖𝒚 +
.
𝟑𝟐
𝟏𝟔

The critical points are where both partial derivatives are zero, i.e. 2𝑥 + 2 = 0, 2𝑦 − 10 = 0,
so the answer is (−𝟏, 𝟓).

𝜕𝑓
2(2𝑥 + 𝑦) − 2(2𝑥 − 𝑦)
4𝑦
(𝑥, 𝑦) =
=
,
(2𝑥 + 𝑦)2
(2𝑥 + 𝑦)2
𝜕𝑥
𝜕𝑓
−1(2𝑥 + 𝑦) − 1(2𝑥 − 𝑦)
−4𝑥
(𝑥, 𝑦) =
=
.
(2𝑥 + 𝑦)2
(2𝑥 + 𝑦)2
𝜕𝑦
This way
𝜕𝑓
4
𝟒 𝜕𝑓
−16
𝟏𝟔
(4,1) =
(4,1) =
=
,
=− .
2
2
(9)
(9)
𝜕𝑥
𝟖𝟏 𝜕𝑦
𝟖𝟏

−7(5𝑥 + 𝑦) − 5(−7𝑥 − 3𝑦)
𝟖𝒚
=
,
2
(5𝑥 + 𝑦)
(𝟓𝒙 + 𝒚)𝟐
−3(5𝑥 + 𝑦) − 1(−7𝑥 − 3𝑦)
−𝟖𝒙
𝑓𝑦 (𝑥, 𝑦) =
=
.
2
(5𝑥 + 𝑦)
(𝟓𝒙 + 𝒚)𝟐
𝑓𝑥 (𝑥, 𝑦) =

(𝑎) 𝑧𝑦𝑥 = 𝑧𝑥𝑦 = 𝟔𝒚,
(𝑏) 𝑧𝑥𝑦𝑥 = (𝑧𝑥𝑦 ) = 𝟎,
𝑥

(𝑐) 𝑧𝑥𝑦𝑦 = (𝑧𝑥𝑦 ) = 𝟔.
𝑦

(I think the exponent is 5)
𝑓(𝑥) = −(5𝑥 + 𝑦)5 ,

𝜕𝑓
= −5 ∙ 5(5𝑥 + 𝑦)4 = −25(5𝑥 + 𝑦)4 ,
𝜕𝑥

𝜕 2𝑓
= −25 ∙ 4(5𝑥 + 𝑦)3 = −𝟏𝟎𝟎(𝟓𝒙 + 𝒚)𝟑 ,
𝜕𝑥𝜕𝑦
𝜕 3𝑓
= −100 ∙ 3 ∙ 5(5𝑥 + 𝑦)2 = −𝟏𝟓𝟎𝟎(𝟓𝒙 + 𝒚)𝟐 ,
𝜕𝑥𝜕𝑦𝜕𝑥
𝜕 3𝑓
𝜕 3𝑓
=
= −𝟏𝟓𝟎𝟎(𝟓𝒙 + 𝒚)𝟐 .
2
𝜕𝑥 𝜕𝑦 𝜕𝑥𝜕𝑦𝜕𝑥

𝑓𝑥 (𝑥, 𝑦) = 𝒚𝒆−𝟐𝒚 ,
𝑓𝑦 (𝑥, 𝑦) = 𝒙𝒆−𝟐𝒚 − 𝟐𝒙𝒚𝒆−𝟐𝒚 ,
𝑓𝑥𝑦 (𝑥, 𝑦) = 𝒆−𝟐𝒚 − 𝟐𝒚𝒆−𝟐𝒚 ,
𝑓𝑦𝑥 (𝑥, ...