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Unsure on how to approach sin^4

  1. Oct 30, 2006 #1
    For d/dx (cos 4x + sin^(4) x)

    is it.

    -sin(4x)(4) + cos^4 (x)??? I was unsure on how to approach sin^4. Thanks.

    Also for d/dx (1)/(16-t^2)^(1/4)

    Is this correct: (-1/4 (16-t^2)^(-5/4))\((16-t^2)^(1/4))^2??

    Thank you.
     
  2. jcsd
  3. Oct 30, 2006 #2
    For (1) [tex] \frac{d}{dx} \cos 4x + \sin^{4} x = -4\sin 4x + 4\sin^{3} x \cos x [/tex]. You have to use both the power rule and the chain rule.

    [tex] \frac{d}{dx} \sin^{4}x, \ u = \sin x [/tex]

    [tex] \frac{d}{du} u^{4} du = 4u^{3} du [/tex]


    For (2) the first half is right. The second half, you have to use the chain rule.
     
    Last edited: Oct 30, 2006
  4. Oct 30, 2006 #3
    Y use the chain rule isnt the second half the bottom of the quotient rule, so it stays like that?/
     
  5. Oct 30, 2006 #4
    [tex] \frac{d}{dt} (16-t^{2})^{-\frac{1}{4}} = -\frac{1}{4}(16-t^{2})^{-\frac{5}{4}}(-2t) [/tex]. So we have used the chain rule.
     
    Last edited: Oct 30, 2006
  6. Oct 30, 2006 #5
    O that goes on the top.. the bottom is correct tho right?
     
  7. Oct 31, 2006 #6

    HallsofIvy

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    quote= helpm3pl3ase]Also for d/dx (1)/(16-t^2)^(1/4)

    Is this correct: (-1/4 (16-t^2)^(-5/4))\((16-t^2)^(1/4))^2??[/quote]
    You can use the quotient rule (with the chain rule):
    [tex]\frac{(d/dx(1))(16-t^2)^{1/4}- (1)(d/dx(16-t^2)^{1/4})}{(16-t^2)^{1/2}}[/tex]
    [tex]= -\frac{(1)(1/4)(16-t^2)^{-3/4}(-2t)}{(16-t^2)^{1/2}}[/tex]
    [tex]= \frac{1}{2}\frac{t}{(16-t^2)^{5/4}}[/tex]

    But it is much easier to write the function as
    [tex](16- t^2)^{-1/4}[/itex]
    and use the chain rule directly:
    [tex]d/dx(16- t^2)^{-1/4}= -1/4(16- t^2)^{-5/4}(-2t)[/tex]
    [tex]= \frac{1}{2}\frac{t}{(16-t^2)^{5/4}}[/tex]
     
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