Calculus I

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  • E2P22
  • April 2, 2013 at 2:52 am #13237

    Ezra Halleck
    Participant

    22. Find the derivative for a function given as a quotient (answer must be a simple fraction and no neg exponents!)

    April 15, 2013 at 1:07 am #16848

    H Pardon
    Member

    This was a difficult problem!!!
    Okay first look at the problem and you will notice a negative exponent -3
    get rid of the negative exponent by making it 1/x^3
    So basically the problem should read xtanx/x2 +1/x^3
    the x in the numerator should cancel with the denominator x^2
    tanx/x + 1/x^3
    From here basically take d/dx (tan/x) + d/dx(1/x^3)

    April 15, 2013 at 1:32 am #16852

    Devindra
    Participant

    This is what I got for my answer
    = tan(x) / ( x^2+ (1/ x^3)) − ((x⋅(2x−3/x^4))⋅tan(x)) / (x^2+(1/x^3))^2 + (x⋅(sec(x))^2 / (x^2+(1/x^3))

    April 15, 2013 at 1:42 am #16855

    H Pardon
    Member

    I got xsec^2x-tanx/x^2 + -3/x^4

    April 15, 2013 at 3:48 am #16871

    Ezra Halleck
    Participant

    First step recommended is to multiply numerator and denominator by x^3 to get
    x^4 tanx/(x^5+1)
    Using quotient rule, we get
    [5x^8tanx-(4x^3tanx +x^4sec^2x)(x^5+1)]/(x^5+1)^2

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