Metamath Proof Explorer


Theorem dvcnsqrt

Description: Derivative of square root function. (Contributed by Brendan Leahy, 18-Dec-2018)

Ref Expression
Hypothesis dvcncxp1.d ⊢ D = ℂ ∖ −∞ 0
Assertion dvcnsqrt ⊢ dx ∈ D x d ℂ x = x ∈ D ⟼ 1 2 ⁢ x

Proof

Step Hyp Ref Expression
1 dvcncxp1.d ⊢ D = ℂ ∖ −∞ 0
2 halfcn ⊢ 1 2 ∈ ℂ
3 1 dvcncxp1 ⊢ 1 2 ∈ ℂ → dx ∈ D x 1 2 d ℂ x = x ∈ D ⟼ 1 2 ⁢ x 1 2 − 1
4 2 3 ax-mp ⊢ dx ∈ D x 1 2 d ℂ x = x ∈ D ⟼ 1 2 ⁢ x 1 2 − 1
5 difss ⊢ ℂ ∖ −∞ 0 ⊆ ℂ
6 1 5 eqsstri ⊢ D ⊆ ℂ
7 6 sseli ⊢ x ∈ D → x ∈ ℂ
8 cxpsqrt ⊢ x ∈ ℂ → x 1 2 = x
9 7 8 syl ⊢ x ∈ D → x 1 2 = x
10 9 mpteq2ia ⊢ x ∈ D ⟼ x 1 2 = x ∈ D ⟼ x
11 10 oveq2i ⊢ dx ∈ D x 1 2 d ℂ x = dx ∈ D x d ℂ x
12 1p0e1 ⊢ 1 + 0 = 1
13 ax-1cn ⊢ 1 ∈ ℂ
14 2halves ⊢ 1 ∈ ℂ → 1 2 + 1 2 = 1
15 13 14 ax-mp ⊢ 1 2 + 1 2 = 1
16 12 15 eqtr4i ⊢ 1 + 0 = 1 2 + 1 2
17 0cn ⊢ 0 ∈ ℂ
18 addsubeq4 ⊢ 1 ∈ ℂ ∧ 0 ∈ ℂ ∧ 1 2 ∈ ℂ ∧ 1 2 ∈ ℂ → 1 + 0 = 1 2 + 1 2 ↔ 1 2 − 1 = 0 − 1 2
19 13 17 2 2 18 mp4an ⊢ 1 + 0 = 1 2 + 1 2 ↔ 1 2 − 1 = 0 − 1 2
20 16 19 mpbi ⊢ 1 2 − 1 = 0 − 1 2
21 df-neg ⊢ − 1 2 = 0 − 1 2
22 20 21 eqtr4i ⊢ 1 2 − 1 = − 1 2
23 22 oveq2i ⊢ x 1 2 − 1 = x − 1 2
24 1 logdmn0 ⊢ x ∈ D → x ≠ 0
25 2 a1i ⊢ x ∈ D → 1 2 ∈ ℂ
26 7 24 25 cxpnegd ⊢ x ∈ D → x − 1 2 = 1 x 1 2
27 23 26 eqtrid ⊢ x ∈ D → x 1 2 − 1 = 1 x 1 2
28 9 oveq2d ⊢ x ∈ D → 1 x 1 2 = 1 x
29 27 28 eqtrd ⊢ x ∈ D → x 1 2 − 1 = 1 x
30 29 oveq2d ⊢ x ∈ D → 1 2 ⁢ x 1 2 − 1 = 1 2 ⁢ 1 x
31 1cnd ⊢ x ∈ D → 1 ∈ ℂ
32 2cnd ⊢ x ∈ D → 2 ∈ ℂ
33 7 sqrtcld ⊢ x ∈ D → x ∈ ℂ
34 2ne0 ⊢ 2 ≠ 0
35 34 a1i ⊢ x ∈ D → 2 ≠ 0
36 7 adantr ⊢ x ∈ D ∧ x = 0 → x ∈ ℂ
37 simpr ⊢ x ∈ D ∧ x = 0 → x = 0
38 36 37 sqr00d ⊢ x ∈ D ∧ x = 0 → x = 0
39 38 ex ⊢ x ∈ D → x = 0 → x = 0
40 39 necon3d ⊢ x ∈ D → x ≠ 0 → x ≠ 0
41 24 40 mpd ⊢ x ∈ D → x ≠ 0
42 31 32 31 33 35 41 divmuldivd ⊢ x ∈ D → 1 2 ⁢ 1 x = 1 ⋅ 1 2 ⁢ x
43 1t1e1 ⊢ 1 ⋅ 1 = 1
44 43 oveq1i ⊢ 1 ⋅ 1 2 ⁢ x = 1 2 ⁢ x
45 42 44 eqtrdi ⊢ x ∈ D → 1 2 ⁢ 1 x = 1 2 ⁢ x
46 30 45 eqtrd ⊢ x ∈ D → 1 2 ⁢ x 1 2 − 1 = 1 2 ⁢ x
47 46 mpteq2ia ⊢ x ∈ D ⟼ 1 2 ⁢ x 1 2 − 1 = x ∈ D ⟼ 1 2 ⁢ x
48 4 11 47 3eqtr3i ⊢ dx ∈ D x d ℂ x = x ∈ D ⟼ 1 2 ⁢ x