M 408D - Cheat Sheet

M 408D - Cheat Sheet - Derivatives 1 x 1 x ln a[xn =...

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Unformatted text preview: Derivatives 1 x 1 x ln a [xn ] = nxn−1 [ln |x|] = [ex ] = ex [ax ] = [ax ] ln a [cos x] = − sin x [csc x] = − csc x cot x [sin x] = cos x [sec x] = sec x tan x [tan x] = sec2 x [cot x] = csc2 x 1 [sin−1 x] = √1−x2 1 [csc−1 x] = − x√x2 −1 1 [cos−1 x] = − √1−x2 1 [sec−1 x] = x√x2 −1 1 [tan−1 x] = 1+x2 x 1 [tan−1 x] = − 1+x2 x [cf ] = cf [f g ] = f g + f g chain rule: [f (g (x))] = f (g (x))g (x) [loga x] = f g f g −f g g2 = Indefinite Integrals integration by parts: un+1 n+1 un du = +C udv = uv − du u , n = −1 v du = ln |u| + C au ln a eu du = eu + C au du = sin udu = − cos u + C tan udu = − ln cos u + C sec udu = ln(sec u + tan u) + C cos udu = sin u + C cot udu = ln sin u + C csc udu = ln(csc u − cot u) + C sec2 udu = tan u + C tan2 udu = tan u − u + C csc2 udu = − cot u + C cot2 udu = − cot u − u + C du u2 +a2 du u2 −a2 = √ du a2 −u2 √ du u2 +a2 1 a tan−1 u a +C = sin−1 u + C a√ = ln u + u2 + a2 + C √ du u u2 −a2 √ du u u2 +a2 = 1 a sec−1 1 = − a ln u a +C √ a+ u2 +a2 u = √ du u2 −a2 1 ln , a > 0, a = 1 u−a u+a = ln u + √ du u a2 −u2 +C 1 2a +C 1 = − a ln √ +C u 2 − a2 + C √ a+ a2 −u2 u +C ...
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This note was uploaded on 09/18/2011 for the course M 408 D taught by Professor Textbookanswers during the Fall '07 term at University of Texas.

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