39 Differentials

Differentials For approximation purposes, we could identify a derivative dy = f 0(x) dx the symbol dx

with

the symbol dy

with

with with

the symbolic product dy · dx with dx the symbolic equation dy · dx with dy = dx

its limiting ratio f (x + h) − f (x) ∆y = , ∆x h h f (x + h) − f (x) f 0(x) · h f (x + h) − f (x) ∼ = f 0(x) · h

Equivalently, we can write f (x + h) ∼ = f (x) + f 0(x) · h, as if the o(h) didn’t exist, and as if the function’s graph were replaced by its tangent line. Example: For the volume V of a cube with side s, V (s) = s3 V (s + h) = (s + h)3 = s3

+ 3s2h

+ 3sh2 + h3

V (s + h) = V (s) + V 0(s)h + o(h) V (s + h) ∼ = V (s) + V 0(s)h representing V (s + h) − V (s) ∼ = V 0(s)h dV ∼ dV ds = ds

f (x + h) ∼ = f (x) + f 0(x) · h 1 1 −1 1 1 ∼ 6 6 + x6 · h (x + h) = x 6 1 1 x6 1 1 ∼ ·h + · (x + h) 6 = x 6 6 x With x = 64 and h = 1 this becomes: √ 6 √ √ 1 64 6 6 ∼ ·1 65 = 64 + · 6 64 √ 1 2 6 ∼ ·1 65 + · = 2 6 64 √ 385 1 6 ∼ = = 2.00521 65 + = 2 192 192 √ (Actually, 6 65 equals 2.00517.)

f (x + h) ∼ = f (x) + f 0(x) · h 1 1 −1 1 1 ∼ 2 2 + x2 · h (x + h) = x 2 1 1 x2 1 1 ∼ ·h + · (x + h) 2 = x 2 2 x With x = 121 and h = 4 this√ becomes: √ √ 1 121 ∼ ·1 125 = 121 + · 2 121 √ 1 11 ∼ ·1 125 + · = 11 2 121 √ 123 1 ∼ = = 11.1818181818 125 + = 11 22 11 √ (Actually, 121 equals 11.18034.)

∼ = f (x) + f 0(x) · h 1 − 1 −1 − 15 ∼ − 15 (x + h) − x 5 ·h = x 5 − 15 1 x − 15 ∼ − 15 ·h − · (x + h) = x 5 x With x = 32 and h = 1 this becomes: − 15 1 32 − 15 − 15 ∼ ·1 − · 33 = 32 5 32 1 1 1 1 ∼ 33− 5 − · 2 ·1 = 2 5 32 1 159 1 − 15 ∼ − = = 0.496875 33 = 2 320 320 − 15 (Actually, 33 equals 0.49693.) f (x + h)

∼ = f (x) cos (x + h) ∼ = cos x

f (x + h)

+ f 0(x) · h

− sin x · h

With x = 60o and h = −3o this becomes: cos 57o cos 57o cos 57o

∼ = cos 60o − sin 60o · (−3o) √ 3π
1 3 ∼ − · − = 2 2 180 ∼ = 0.5000 + 0.0453 = 0.5453

(Actually, cos 57o equals 0.5446.)

∼ = f (x) + f 0(x) · h 1 ∼ ·h ln (x + h) = ln x + x 1 With x = 1 and h = 10 this becomes: f (x + h)

ln 1.1 ln 1.1

∼ = ln 1 + 1 · (0.1) ∼ = 0.0 + 0.1 = 0.1

(Actually, ln 1.1 equals 0.09531.)

f (x + h) ∼ = f (x) + f 0(x) · h ∼ + ex · h e(x+h) = ex With x = 0.0 and h = 0.1 this becomes: e0.1 e0.1

∼ = e0.0 ∼ = 1.0

+ e0.0 · (0.1) + 1.0 · (0.1) = 1.1

(Actually, e1.1 equals 1.10517.)