Web-matematik (k)’ = 0 (x)’ = 1 (ax + b)’ = a (x2)’ = 2x (x3)’ = 3x2 ( 1x )’ = − 1x2 (√x)’ = 12√x (ax)’ = ax ⋅ ln(a) Differentialregning (9) (ex)’ = ex (ekx)’ = k ⋅ ekx ln’(x) = 1x Differentialregning (8) log’(x) = 1x ⋅ ln(10) Differentialregning (12) (xa)’ = a ⋅ xa − 1 (xx)’ = xx ⋅ (ln(x) + 1) Differentialregning (13) |
( 1f (x) )’ = − f ’ (x)f 2 (x) ⇒ ( 1x )’ = − 1x2 1. Bevis: ( 1x )’ = ( x0x )’ = ( x0 − 11 )’ = − x−21 = − x−2 ⋅ x21 ⋅ x2 = − x−2 + 2x2 = − x0x2 = − 1x2 2. Bevis: (1)’ = ( x ⋅ 1x )’ = (x)’ ⋅ 1x + x ⋅ ( 1x )’ = 1 ⋅ 1x + x ⋅ ( 1x )’ = 0 ( 1x )’ = −1xx = − 1x2 |
torsdag den 11. juni 2015
Differentialkvotienter
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