Calculus Reference Basic Differentiation Rules ππ [ππππ] = ππππβ² ππππ ππ [π’π’π’π’] = π’π’π£π£ β² + π£π£π’π’β² (product rule) ππππ ππ [ππ] = 0 (constant) ππππ ππ [π₯π₯] = 1 ππππ ππ π’π’β² [ln π’π’] = (natural log rule) ππππ π’π’ ππ [ππππ π₯π₯ ] = ππ (ln ππ) ππ π₯π₯ for b > 0 ππππ ππ [π’π’ ± π£π£] = π’π’β² ± π£π£β² ππππ ππ π’π’ π£π£π’π’β² β π’π’π’π’β² οΏ½ οΏ½= (quotient rule) ππππ π£π£ π£π£ 2 ππ ππ [π’π’ ] = ππ π’π’ππβ1 π’π’β² ππππ ππ π’π’ β² [ |π’π’| ] = π’π’ for u β 0 |π’π’| ππππ ππ π’π’ [ππ ] = ππ π’π’ π’π’β² ππππ ππ π₯π₯ [π’π’ ] = ux ln π’π’ ππππ Basic Integration Formulas οΏ½ ππ ππ(π’π’)ππππ = ππ οΏ½ ππ(π’π’)ππππ οΏ½[ππ(π’π’) ± ππ(π’π’)] ππππ = οΏ½ ππ(π’π’)ππππ ± οΏ½ ππ(π’π’)ππππ οΏ½ ππππ = π’π’ + πΆπΆ οΏ½ οΏ½ π’π’ππ ππππ = ππππ = ln|π’π’| + πΆπΆ π’π’ π’π’ππ+1 + πΆπΆ, for n β β1 ππ + 1 οΏ½ ππ π’π’ ππππ = ππ π’π’ + πΆπΆ Properties of Exponents and Logarithms Exponent 0 < ππ β 1, 0 < ππ β 1, π’π’ > 0, π£π£ > 0 Logarithm Example Rule Rule 22 24 = 22+4 = 26 ππ π₯π₯ ππ π¦π¦ = ππ π₯π₯+π¦π¦ log ππ (π’π’π’π’) = log ππ π’π’ + log ππ π£π£ (32 )3 = 36 (ππ π₯π₯ )π¦π¦ = ππ π₯π₯π₯π₯ log ππ (π’π’ππ ) = ππ log ππ π’π’ log 3 (94 ) = 4 log 3 9 = 4 β 2 = 8 81 = 8 ππ1 = ππ log ππ (ππ) = 1 log 5 (5) = 1 78 = 78β6 = 72 = 49 76 50 = 1 1 = 4β2 42 (20)3 = (4 β 5)3 = 43 53 6 = log 2 26 Jennifer B ππ π₯π₯ = ππ π₯π₯βπ¦π¦ ππ π¦π¦ ππ ππ = 1 1 = ππ βπ₯π₯ ππ π₯π₯ (ππππ) π₯π₯ = ππ π₯π₯ ππ π₯π₯ π₯π₯ = log ππ ππ π₯π₯ π’π’ log ππ οΏ½ οΏ½ = log ππ π’π’ β log ππ π£π£ π£π£ log ππ (1) = 0 1 log ππ οΏ½ οΏ½ = β log ππ π’π’ π’π’ log ππ π’π’ log ππ (π’π’) = log ππ ππ π’π’ = ππ log ππ π’π’ Example log 2 (4 β 16) = log 2 4 + log 2 16 =2+4=6 8 log 2 οΏ½ οΏ½ = log 2 8 β log 2 4 4 =3β2=1 log 4 (1) = 0 1 log 2 οΏ½ οΏ½ = β log 2 32 = β5 32 log 2 16 4 log 4 16 = = =2 log 2 4 2 8 = 5log 5 8 May 2010
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