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Homework Help: Derrivitive problem

  1. Oct 12, 2004 #1
    Well I was doing my calculus homework today and I got stuck one some problems. Well actually I just wanted to check if I did them right because I'm not entirely sure. Anybody care to check? :D

    1) y = sin (9[x^(1/2)] + sin [x^(1/2)]
    --For this problem, I just used the chain rule for each part. For my answer I got [cos(x^91/2)]/[x^(1/2)]. I wasn't sure if I could do the chain rule for both of them and just add them. Was that okay?

    2) y = 4sec^2 x(tan x)
    --In this problem, I used the product rule then the chain rule inside. I think I did it right but the 4 kind of threw me off. My final answer was: 4sec^2x (sec^2x + 2tan^2x).

    Thanks! Anythign helps really... :D
     
  2. jcsd
  3. Oct 12, 2004 #2

    NateTG

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    It's much easier to help if you show work. That said:

    1.The way you've written it, the parens don't balance, but in general, the derivative of a (finite) sum is the sum of the derivatives. Also your answer doesn't look right to me I think you did some bad algebra pulling things out of the trig functions.

    2. It seems like there should be an extra 2 from the [tex]\sec^{x}[/tex] in the derivative that you're missing.
     
  4. Oct 12, 2004 #3
    HAha yeah it would be easier that way wouldn't it? Okay here's my work, that way I can pin point wher eI made a mistake, apparently. :D

    1) y = sin x1/2 + sin x1/2
    y' = (cos x1/2)(1/2x-1/2) + (cos x1/2)(1/2x-1/2)
    y' = (2cosx1/2)/(2x1/2 )
    y' = cos x1/2 /x1/2

    For the second problem...
    y = 4sec2x(tan x)
    y' = tan x(8 sec x(sec x * tan x)) + 4(sec x)2(sec2x
    y' = tan2 x(2)) + 4(sec x)2(sec2x
    y' = 4 sec2x[sec2 x +2 tan2 x]

    I hope that helps as much as it may help me. :D
     
  5. Oct 13, 2004 #4

    NateTG

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    The way you originally posted it, it looked more like:
    [tex]y=\sin(9x^{\frac{1}{2}}) + \sin(x^{\frac{1}{2}})[/tex]
    which would lead to a different derivative. The stuff above all looks good.
    Also looks good. Here's a quick check:
    [tex]y=4 \sec^2x\tanx=4\frac{\sin x}{\cos^3 x}=4 \sin x \cos^{-3} x[/tex]
    [tex]y'=4(\cos x \cos^{-3} x + \sin x (-3) \cos^{-4} x(-\sin x))[/tex]
    [tex]y'=4(\cos^{-2} x + 3 \sin^2 \cos^{-4})[/tex]
    [tex]y'=4 \sec^2x(1+3 \tan^2 x)[/tex]
    and
    [tex]1+\tan^2=\sec^2[/tex]
    so
    [tex]y'=4 \sec^2x(\sec^2 x + 2 \tan^2 x)[/tex]
     
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