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Trigonometric derivatives

  1. Mar 30, 2007 #1
    How do I get the derivative of
    y=tan4x
    The answer in the back of the text book
    says (4/cos^2(4X))
     
  2. jcsd
  3. Mar 30, 2007 #2

    mjsd

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    remember tan x can be written in terms of sin and cos
     
  4. Mar 30, 2007 #3
    yes i know that tanx= sinx/cosx
    y=tan4x -> y=sin4x/cos4x
    Then what?
     
  5. Mar 30, 2007 #4

    Gib Z

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    Also use the chain rule, u=4x.
     
  6. Mar 30, 2007 #5
    y=sin4x/cos4x
    u=4x
    u'=4
    y=sinu/cosu
    y=cosu/-sinu
    y=4cosu/-4sinu
    y=4cos4x/-4sin4x
    What am I doing wrong?
     
  7. Mar 30, 2007 #6

    Gib Z

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    Ahh no have you learned the quotient rule? Ask your teacher about that.

    If you have a function f(x)=u/v then the derivative is f'(x)=[tex]\frac{u'v-v'u}{v^2}[/tex]. So let u=sin x and v = cos x for this, you can find the derivative of tan x is (sec x)^2, then after that use the chain rule.
     
  8. Mar 30, 2007 #7
    Using the quotient rule this is what i get.
    u=sin4x
    u'=4cos4x
    v=cos4x
    v'=-4sin4x

    (-4sinX)(cos4X)--(4sinX)(sin4X)/(cos4X)^2
     
  9. Mar 30, 2007 #8

    Gib Z

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    No no, forget about the 4x bit for now, just use the quotient rule with u=sin x and v=cos x to find the derivative of tan x.

    One you know that, THEN use the chain rule. Do you know what the chain rule is?
     
  10. Mar 30, 2007 #9
  11. Mar 30, 2007 #10
    Do you mind showing me the whole working out yourself?
    That would help me alot.
    Thanks
     
  12. Mar 30, 2007 #11
    The video was more than sufficient.

    If you still are unable to complete the problem, seek help from your teacher, peers, or get a tutor.
     
  13. Mar 30, 2007 #12
    [tex]y=tan4x ; u=4x[/tex]
    [tex]\frac{dy}{dx}=\frac{d(tanu)}{du}.\frac{du}{dx}[/tex]
    [tex]\Rightarrow \ \frac{dy}{dx}= sec^2u. 4 [/tex]
    [tex]\Rightarrow \ \frac{dy}{dx}=4sec^{2}4x[/tex]
     
    Last edited: Mar 30, 2007
  14. Mar 30, 2007 #13
    We havent learnt the other ratios yet.
    The answer i need to get is in this form
    (4/cos^2(4X))
    I've finally figured out how to do this i just don't understand how
    (4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
    What happens to the cos and sin squared in the numerator?
     
  15. Mar 30, 2007 #14

    hotvette

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    Hint: isn't there a trig identity relating sin2 and cos2? :smile:
     
  16. Mar 30, 2007 #15
    too easy
    thanks guys
     
  17. Mar 30, 2007 #16
    i thought you would know [itex] \ sec^2x=\frac{1}{cos^2x} [/itex] :uhh:
     
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