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Chain rule app.

  1. Dec 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [tex]y=\sqrt{x+f(x^2-1)}[/tex]. Find [tex]\frac{dy}{dx}[/tex] when x=3, given that [tex]f(8)=0[/tex] and [tex]f'(8)=3[/tex].


    I missed the lesson for this and am lost on what to do. some guidence would be appreciated. i dont understand the [tex] f(x^2-1) [/tex] inside the bracket.
     
    Last edited: Dec 6, 2007
  2. jcsd
  3. Dec 6, 2007 #2
    i spent like an hour trying to figure this out. (need to not miss class!)

    my attempt at a solution:

    [tex]\frac{dy}{dx}=\frac{1+f'(x^2-1)}{2\sqrt{x+f(x^2-1)}}[/tex]


    [tex]\frac{dy}{dx}=\frac{1+f'(3^2-1))}{2\sqrt{3+f(3^2-1)}}[/tex]


    [tex]\frac{dy}{dx}=\frac{1+f'(3^2-1))}{2\sqrt{3+f(3^2-1)}}[/tex]

    [tex]\frac{dy}{dx}=\frac{1+f'(8))}{2\sqrt{3+f(8)}}[/tex]

    [tex]\frac{dy}{dx}=\frac{1+3}{2\sqrt{3+0}}[/tex]

    [tex]\frac{dy}{dx}=\frac{4}{2\sqrt{3}}[/tex]

    [tex]\frac{dy}{dx}=\frac{2}{\sqrt{3}}[/tex]

    this this anywhere near correct? can someone inform me where i went wrong? right of the bat? inbetween??
     
  4. Dec 6, 2007 #3
    You're very close. However, you forgot to apply the chain rule when you took the derivative of [tex]f(x^2 - 1)[/tex]. That, too, is a composition of function and so you must also apply the chain rule when taking its derivative.
     
  5. Dec 6, 2007 #4
    am not sure i understand but...would its derivative be[tex]\frac{1}{f(x^2-1)}[/tex]? or could t be [tex]f'(x^2-1)(2x)[/tex]
     
    Last edited: Dec 6, 2007
  6. Dec 6, 2007 #5
    Not quite. Remember that the chain rule says

    [tex]\frac{d}{dx}f(g(x))=f'(g(x))g'(x)[/tex]​

    When you have

    [tex]f(x^2 - 1)[/tex]​

    They're saying that, rather than plugging in the usual x in your variable, you plug in [tex]x^2 - 1[/tex]. For example, let [tex]f(t) = t^2[/tex]. Then, [tex]f(x^2) = (x^2)^2 = t^4[/tex]. So, you have a composition of functions, where, using the notation of the chain rule as stated above,

    [tex]f(x) = f(x)[/tex]

    [tex]g(x) = x^2 - 1[/tex]​

    So, thinking about the chain rule, what would the derivative with respect to x be of [tex]f(x^2 - 1)[/tex]?
     
    Last edited: Dec 6, 2007
  7. Dec 6, 2007 #6
    Yes, the correct answer is [tex]f'(x^2-1)(2x)[/tex].
     
  8. Dec 6, 2007 #7
    so new solution is:

    [tex]\frac{dy}{dx}=\frac{1+f'(x^2-1)(2x)}{2\sqrt{x+f(x^2-1)}}[/tex]

    [tex]\frac{dy}{dx}=\frac{1+f'(8)(2(3))}{2\sqrt{3+f(8)}}[/tex]

    [tex]\frac{dy}{dx}=\frac{1+18}{2\sqrt{3}}[/tex]

    [tex]\frac{dy}{dx}=\frac{19}{2\sqrt{3}}[/tex]
     
  9. Dec 6, 2007 #8
    Yup. That's right.
     
  10. Dec 6, 2007 #9

    great. thanks a lot man. really starting to understand.

    another problem. i have soloved it, but need to see if i have done it correctly.

    Q-skydiver jumps from plane at 3000m. distance fallen in meters after t seconds is:

    [tex] s=5t^2[/tex]

    During fall, experiences air pressure [tex]p[/tex] that will cause his ears to pop if the rate of change of pressure [tex]\frac{dp}{dt}[/tex] exceeds 2 pressure units/s. suppose that the rate of change of pressure with respect to distance fallen in metres is 0.075 pressure units/m. what time will the sky divers ears pop? At what height will this occur?

    Solution attempt:
    A)time ears pop

    [tex]\frac{dp}{dt}=\frac{dp}{ds}*\frac{ds}{dt}[/tex]

    [tex]\frac{dp}{dt}=2[/tex]

    [tex]\frac{dp}{ds}=0.075[/tex]

    [tex]\frac{ds}{dt}= 10t[/tex]

    [tex]2=0.075*10t[/tex]

    [tex]t=2.7[/tex]

    b)height of poping

    [tex]s=5(2.7)^2[/tex]
    [tex]s=35.6[/tex]
    [tex]3000-35.6=2964.4[/tex]
     
  11. Dec 6, 2007 #10
    Yes, that looks right too.
     
  12. Dec 6, 2007 #11
    cool. that's 2 out of 4 on the worksheet. onto number three.

    suppose that f and g are functions such that:

    [tex]f(1)=-\frac{1}{2}[/tex] , [tex]f'(1)=-\frac{2}{3}[/tex] , [tex]g(2)=1[/tex] and [tex]g'(2)=3[/tex]

    find [tex]h'(2)[/tex] where [tex]h[/tex] is the composite fuction [tex]h(x)=f(g(x))[/tex].

    so in this to get [tex]h'(2)[/tex] i need to just take the derivative [tex]f'(1)[/tex] and multiply it by the derivative [tex]g'(2)[/tex] ?

    [tex]-\frac{2}{3} * 3 [/tex].

    so [tex]h'(2)[/tex] is equal to -2?

    it seems wrong to me because i have done it too simply. did i completely miss the mark? or partialy?
     
  13. Dec 6, 2007 #12
    I think you understand the chain rule, so you should have a good idea whether that is correct or not.
     
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