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Homework Help: Inverse Trig Derivatives

  1. Jun 13, 2005 #1
    hi,
    my problem is i have equations like determine the derivative of:

    a. arcsin 1/x
    b. arcsin x^(1/2)
    c. arcsin x^(-1/2)
    etc etc.

    and i have the formulas for the trigs such as:

    d/dx arcsinf(x) = 1/(1-f(x)^2)^1/2 x f'(x) and i have the arcos and asrctan one.

    but when i put f(x) into the equations eg:

    arcsin^(1/2):

    1/ (1-x^1/2)^1/2 x 1/2x^-1/2

    i can never get the answer i know im multiplying wrong and possibly using the powers wrong because and example the teacher used was:

    y = arctan 3/x^2 find y'

    f(x) = 3x^-2
    f'(x) = -6x^-3

    y'= 1/1+(3x^-2)^2 x (-6x^-3)

    therefore: y' = -6x/ x^4 + 9

    how did he get the final answer from y'= 1/1+(3x^-2)^2 x (-6x^-3)??? im totaly confused ive been trying to work it out for days any help would be appreciated thanks
     
  2. jcsd
  3. Jun 13, 2005 #2
    [tex] y = arctan\left(\frac{3}{x^2}\right) [/tex]

    [tex] \frac{d}{dx} arctan(x) = \frac{1}{1+x^2}} [/tex]

    By chain rule

    [tex] \frac{d}{dx} arctan(f(x)) = \frac{f'(x)}{1+f(x)^2}} [/tex]

    [tex] f(x) = \frac{3}{x^2} \Rightarrow \frac{d}{dx} f(x) = \frac{-6}{x^3}[/tex]

    [tex] \frac{d}{dx} arctan\left(\frac{3}{x^2}\right) = \frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)} [/tex] This is a correct answer, but can be simplified to:

    [tex] \frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4+9}{x}} = \frac{-6x}{x^4+9}[/tex]
     
  4. Jun 13, 2005 #3
    i dont understand why it become x^4 here? i understand the whole thing up to this point?
     
  5. Jun 13, 2005 #4
    a+(b/c)
    = (ac/c) + (b/c) .... (since c/c = 1)
    = (ac+b)/c

    -- AI
     
  6. Jun 13, 2005 #5
    [tex] \frac{d}{dx} arctan\left(\frac{3}{x^2}\right) = \frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)} [/tex]

    [tex] \frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)} \times \frac{x^3}{x^3} = \frac{\frac{-6}{x^3}x^3}{(1+\left(\frac{9}{x^4}\right))x^3} = \frac{-6}{x^3+\frac{9}{x}}[/tex]

    Then make a common denominator for the bottom, [itex] x^3 = x^4/x [/tex]

    [tex] \frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4}{x} + \frac{9}{x}}. [/tex]

    Combine fractiosn with a common denominator.

    [tex] \frac{-6}{\frac{x^4+9}{x}} [/tex]

    Dividing is the same as multiplying by the reciprocal.

    [tex] \frac{-6}{\frac{x^4+9}{x}} = -6 \times \frac{x}{x^4+9} [/tex] and you get the final result. This is basic algebra.
     
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