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Math Help - Help with taking the derivative of any of these inverse tangent functions?

  1. #1
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    Exclamation Help with taking the derivative of any of these inverse tangent functions?

    PS I KNOW THE FORMULAS I'M JUST NEW TO THIS AND STILL DON'T QUITE UNDERSTAND IT ALL

    1. g(x) = 3arccos(x/2)

    I got:
    -3/[2*sqrt(4-x^2)]

    The answer is:
    As above without the '2'

    2. f(x) = arctan (x/a)

    I got:
    [1/a] / [1 + x^2/a^2]

    The answer is:
    a / [a^2+x^2]

    3. g(x) = arcsin (3x) / x
    NO IDEA
    The answer is:
    [3x - sqrt(1-9x^2)*arcsin(3x)] / [x^2*sqrt(1-9x^2)]

    4. h(t) = sin[arccos(t)]

    The answer is:
    -t/sqrt(1-t^2)

    5. y = x*arccos(x) - sqrt(1-x^2)

    The answer is:
    arccos(x)

    6. y = (1/2) * [ (1/2)*ln{x+1/X-1} + arctan(x) ]

    The answer is:
    1/[1-x^4]
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  2. #2
    MHF Contributor

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    Quote Originally Posted by dillonmhudson View Post
    PS I KNOW THE FORMULAS I'M JUST NEW TO THIS AND STILL DON'T QUITE UNDERSTAND IT ALL

    1. g(x) = 3arccos(x/2)

    I got:
    -3/[2*sqrt(4-x^2)]

    The answer is:
    As above without the '2'
    The derivative of arccos(x) is -1/\sqrt{1- x^2} so the derivative of g is \frac{-3}{\sqrt{1-(x/2)^2}}(\frac{1}{2})= \frac{-3}{2\sqrt{1-(x/2)^2}}. Now take that "2" inside the squareroot to get \frac{-3}{\sqrt{4- 4(x/2)^2}}= \frac{-3}{\sqrt{4- x^2}}.

    2. f(x) = arctan (x/a)

    I got:
    [1/a] / [1 + x^2/a^2]

    The answer is:
    a / [a^2+x^2]
    Multiply both numerator and denominator of your answer by a^2.

    3. g(x) = arcsin (3x) / x
    NO IDEA
    The answer is:
    [3x - sqrt(1-9x^2)*arcsin(3x)] / [x^2*sqrt(1-9x^2)]
    Use the quotient rule: (arcsin(3x)/x)'= [x(arcsin(3x))'- arcsin(3x)]/x^2 = \frac{\frac{3x}{\sqrt{1-9x^2}}- arcsin(3x)}{x^2}.

    4. h(t) = sin[arccos(t)]

    The answer is:
    -t/sqrt(1-t^2)
    The best way to do this is to note that is y= arccos(t), which is the same as t= cos(y), then we can imagine a right triangle with angle y, hypotenuse of length 1 and "opposite side" t. By the Pythagorean theorem, the "near side" has length \sqrt{1- t^2} and sin(y)= sin(arccos(t))= \sqrt{1- t^2}/1. So h(t)= \sqrt{1- t^2}= (1- t^2)^{1/2}. Use the chain rule to differentiate that: [(1-t^2)^{1/2}= (1/2)(1-t^2)^{-1/2}(-2t)= \frac{-t}{\sqrt{1- t^2}}

    5. y = x*arccos(x) - sqrt(1-x^2)

    The answer is:
    arccos(x)
    Use the product rule for x arccos(x): (x)' arccos(x)+ x(arccos(x))'= arccos(x)- \frac{x}{\sqrt{1-x^2}} while the derivative of -\sqrt{1- x^2} we just did above- its derivative is \frac{x}{\sqrt{1- x^2}}. You see that cancels the previous term, leaving only arccos(x).

    6. y = (1/2) * [ (1/2)*ln{x+1/X-1} + arctan(x) ]

    The answer is:
    1/[1-x^4]
    The first thing I would do is write ln((x+1)/(x-1)) as ln(x+1)- ln(x-1) so its derivative is \frac{1}{x+1}- \frac{1}{x-1}. Of course the derivative of arctan(x) is \frac{1}{1+ x^2}.

    y'= \frac{1}{2(x+1)}- \frac{1}{2(x-1)}+ \frac{1}{1+x^2}. (I've dropped the 1/2 multiplying the whole thing so I won't have to carry it along.) Add those fractions by getting the common denominator 2(x+1)(x-1)(x^2+1)= 2(x^2-1)(x^2+1)= 2(x^4- 1)
    \frac{(x-1)(x^2+1)}{2(x^4-1)}- \frac{(x+1)(x^2+1)}{2(x^4-1)}+ \frac{2(x-1)(x+1)}{2(x^4-1)} = \frac{x^3- x^2+ x- 1- x^3- x^2- x- 1+ 2x^2- 2}{2(x^4-1)} = \frac{-2}{x^4-1}.
    Putting back the 1/2 I dropped gives \frac{-1}{x^4-1}= \frac{1}{1- x^4}.
    Last edited by HallsofIvy; March 29th 2009 at 05:45 PM.
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