1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Inverse Trig Derivatives

  1. Jun 13, 2005 #1
    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:


    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
    = (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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook