шамин с сдо
.pdf
f
x
dy = df = f′(x)Δx = f′(x)dx.
dy = f′(x)dx.
x
f′(x) = df(x) . dx
dx
f(x) = x2 x = 6 dx = 3
y = 2x x + o(Δx),
dy = 3xdx.
x = 5 dx = 3
dy = 2 · 6 · 3 = 36.
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y = x2 |
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y= 12 x − 36 |
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x = 6 |
dx = 3 |
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f(x) = x3 |
x = 5 |
dx = 0.1 |
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y = 3x2 x + o(Δx), |
dy = 3x2dx.
x = 5 dx = 0.1
dy = 3 · 52 · 0.1 = 7.5.
o(Δx) |
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√ |
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15 |
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f(x) = √ |
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x = 15 |
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x0 = 16 |
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x0 = 16 |
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x = −1 |
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dy = |
dx = |
· (−1) = −0.125. |
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2√ |
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2 · 4 |
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x0 |
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dy ≈ f(x0 + x) − f(x0) |
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f(x0 + |
x) ≈ f(x0) + dy. |
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√√
15 ≈ 16 − 0.125 = 3.875.
√
15 ≈ 3.873
(√ |
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1 |
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2√x |
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f(x0 + x) ≈ f(x0) + f′(x0) · x. |
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sin 35◦ |
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sin 30◦ cos 30◦ |
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√ |
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π |
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π |
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sin 35◦ ≈ sin 30◦ + cos 30◦ · 5 |
= 0.5 + |
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5 |
= 0.5756. |
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180 |
2 |
180 |
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sin 35◦ = 0.5736
H(x) = f(g(x)),
y = g(x) z = f(y) g(x)
f(y)
g(x)
xf(y)
y = g(x) |
H(x) |
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x |
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H′(x) = f′(y)g′(x). |
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f(y) |
y |
z = f′(y)Δy + δ(Δy)Δy.
δ(Δy) → 0 y → 0
δ(0) = 0
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g(x) |
z |
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f(y) |
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z |
= f′(y) |
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+ δ(Δy) |
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x → 0
H′(x) = f′(y)g′(x) + lim δ(Δy)g′(x) = f′(y)g′(x).
y→0
h(x) = sin x2
x (a, b) y = f(x)
h(x) = f(g(x)), f(y) = sin y, g(x) = x2.
f′(y) = cos y, g′(x) = 2x,
h′(x) = 2x · cos x2.
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(a, b) |
f(x) |
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f′(x) = 0 |
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6 |
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f−1(y) |
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f−1(y) ′ = |
1 |
, y = f(x). |
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f′(x) |
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f−1(y)
f(x) |
x0 (a, b) |
xx + x (a, b)
y |
f(x) |
y
y = [f′(x0) + ε(Δx)]Δx.
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f(x) |
x |
y |
y |
x = ϕ(Δy) |
ϕ(t) |
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0 |
ϕ(0) = 0 |
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y = [f′(x0) + ε(ϕ(Δy))]Δx.
ε(ϕ(Δy)) → 0 |
y → 0 |
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x |
y |
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1 |
1 |
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= |
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→ |
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, y → 0. |
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f′(x0) + ε(ϕ(Δy)) |
f′(x0) |
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f(x) = x3
f−1(y) = √y
3
(0, ∞)
(f−1
x = y1/3
x3
y
(y))′ = |
1 |
= |
1 |
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f′(x) |
3x2 |
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( |
√x)′ |
= 3y2/3 . |
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1 |
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√ |
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0 |
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3 |
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x = ϕ(t), y = ψ(t),
t (t0, t1) y = f(x)
ϕ(t)
(t0, t1)
ϕ(t) ψ(t)
y = f(x)
y = f(x) = ψ(t) = ψ(ϕ−1(x)).
f(x) |
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f′(x) = ψ′(t) |
1 |
= |
ψ′(t) |
. |
ϕ′(t) |
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ϕ′(t) |
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y = f(x)
x = sin(t), y = cos(t),
t (−1, 1)
f′(x) = − sincosxt = − tg t.
y = f(x)
F (x, y) = 0.
y = f(x)
F (x, f(x)) = 0, x (x1, x2).
F
x
yx
y′
x
x
dy dx
yx + ey = 10.
x
y′x + y + eyy′ = 0.
y′ = − y . x + ey
f
f′(x) |
x |
f′(x)
f′′(x)
n
n
fx
(n − 1)
f(n)(x) = (f(n−1)(x))′.
n
f(x) g(x)
n |
I |
n
(fg)(n) = f(n)g + Cn1f(n−1)
Ck |
= |
n! |
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n |
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k!(n−k)! |
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n = 1
(fg)(n+1) =
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X |
g′ + |
· · · |
+ fg(n) = Ckf(n−k)g(k), |
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k=0 |
n |
n+1 |
n!′
X
Cnkf(n−k)g(k) =
k=0
n
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X
=Cnk f(n−k+1)g(k) + f(n−k)g(k+1) =
k=0 |
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n |
n+1 |
X |
X |
=Cnkf(n+1−k)g(k) + Cnj−1f(n+1−j)g(j) =
k=0 |
j=1 |
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n |
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Cn0f(n+1)g + |
(Cnk + Cnk−1)f(n+1−k)g(k) + Cnnfg(n+1). |
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k=1 |
Cnk + Cnk−1 = Cnk+1
n + 1
(sin x cos x)′′
(sin x cos x)′′ = (sin x)′′ cos x + 2(sin x)′(cos x)′ + sin x(cos x)′′ =
= − sin x cos x − 2 cos x sin x − sin x cos x = −4 sin x cos x.