( xa ) ' = a ⋅ xa − 1 ( 1f ( x ) ) ' = − f ' ( x )f 2 ( x ) ⇒ ( 1x ) ' = − 1x2 ( ekx ) ' = ( kx ) '⋅ ( ekx ) ' = k ⋅ ekx ( √ x ) ' = 12√ x ln ' ( x ) = 1x | ∫ xn dx = 1n + 1 ⋅ xn + 1 ∫ 1x dx = ln ( | x | ) ∫ ax dx = a ⋅ ln ( x ) + k, x > 0 ∫ ax dx = axln ( a ) ∫ ekx dx = 1k ⋅ ekx ∫ √ x dx = 23 ⋅ x1,5 = 23 ⋅ ( √ x )3 |
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