unit 5 formula review s · PDF file · 2016-01-04Log/Exponent Properties:...

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Transcript of unit 5 formula review s · PDF file · 2016-01-04Log/Exponent Properties:...

  • Log/Exponent Properties: ln(1) = 0 ln(e) = 1 ln(an) = n*ln(a) ln(ab) = ln(a) + ln(b)

    baba lnlnln =

    Exponent Properties: ea * eb = ea+b (ea)b = eab e0 = 1

    Log Differentiation steps: 1) Take ln of both sides. 2) Expand right side. 3) Find derivative 4) Solve for dy/dx Evaluate derivative of inverse: (find !! !() 1.Set f(x) = a and solve for x (guess and check) 2. Find f (x) 3. Plug in x value from step #1 into f (x). 4. Flip value.

    Log Derivatives: Exponential Derivatives ddxln | u |= u '

    u

    ddxeu = eu u '

    uu

    au

    dxd

    a'*

    ln1log =

    ddxau = lnaau *u '

    Trig Derivatives: d

    dxsinu = cosu*u '

    ddxtanu = sec2 u*u '

    ddxsecu = secu tanu*u '

    ddxcosu = sinu*u '

    ddxcotu = csc2 u*u '

    ddxcscu = cscucotu*u '

    Inverse Trig Derivatives: ddxarcsinu = u '

    1u2 ddxarctanu = u '

    1+u2 ddxarcsecu = u '

    u u2 1

    ddxarccosu = u '

    1u2 ddxarccotu = u '

    1+u2

    ddxarccscu = u '

    u u2 1

    Integral Formulas: Power Rule:

    un du = un+1

    n+1+C

    Log Rule: 1udu = ln | u |+C

    Exponential Rule: (Base e) dueu = e

    u + C

    Exponential Rule (base other than e)

    au du = au

    lna+C

    *Note: lna is a constant*

    Trig Integrals: sinudu = cosu+C cosudu = sinu+C sec2 udu = tanu+C secu tanudu = secu+C csc2 udu = cotu+C cscucotudu = cscu+C

    C + |cosu|-lntan = udu Cuudu += sinlncot

    udusec = ln|sec u + tan u| + C

    uducsc = -ln|csc u + cot u| + C

    Inverse Trig Integrals:

    +=

    Cau

    uadu arcsin

    22 +=+ Ca

    uaua

    du arctan122

    +=

    Cau

    aauudu ||secarc1

    22

    !"#! ! = and log! ! =

    log! = ln ln

    Interest Formulas

    = !1 +!(!")

    A = Pert