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Can you Differentiate tan2 theta !? watch

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    Parametric Equation

    x= tan theta, y= tan2theta

    Find the value if dy/dx when theta= pi/6

    I differentiated tan theta got 1/cos theta which is sec? but stuck on the next part

    Thx ..
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    Try the chain rule
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    let 2theta = x
    then differentiate tan x and then sub in the thetas in at the end.
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    (Original post by mufas)
    Parametric Equation

    x= tan theta, y= tan2theta

    Find the value if dy/dx when theta= pi/6

    I differentiated tan theta got 1/cos theta which is sec? but stuck on the next part

    Thx ..
    Are you asking about tan^2(\theta) or tan(2\theta)?
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    (Original post by Ben121)
    Try the chain rule
    got 2sec^2x but still dont work
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    (Original post by EEngWillow)
    Are you asking about or ?
    this one
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    (Original post by mufas)
    this one
    In that case use the chain rule with u = 2theta as suggested. Post your working if you got the wrong answer and we can help you further.

    EDIT: Just saw your post above. The argument inside your trig function should still be 2theta. Otherwise you've done it correctly.
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    (Original post by EEngWillow)
    In that case use the chain rule with u = 2theta as suggested. Post your working if you got the wrong answer and we can help you further.

    EDIT: Just saw your post above. The argument inside your trig function should still be 2theta. Otherwise you've done it correctly.
    Again I got 2/cos2theta

    so when (2/cos2theta) / (1/cos theta )

    = 2cos theta / cos 2 theta

    when i sub pi/6 into the top one i get decimals , but the answer is 6!
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    (Original post by mufas)
    Again I got 2/cos2theta

    so when (2/cos2theta) / (1/cos theta )

    = 2cos theta / cos 2 theta

    when i sub pi/6 into the top one i get decimals , but the answer is 6!
    Ah, I've seen your problem. Silly me. The derivitive of tan(theta) isn't sec(theta). Try doing that one again.
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    (Original post by EEngWillow)
    Ah, I've seen your problem. Silly me. The derivitive of tan(theta) isn't sec(theta). Try doing that one again.
    checked again same ans whats wrong? :confused:

    its 2/cos2theta ?
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    try differentiating tan with the quotient rule, remembering tanx = (sinx)/(cosx) and see what the standard result for differentiating tan is
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    (Original post by RK92)
    try differentiating tan with the quotient rule, remembering tanx = (sinx)/(cosx) and see what the standard result for differentiating tan is
    its tan 2 theta not just tan theta
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    Its been 40 min still no solution !! I'm still trying!
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    (Original post by mufas)
    checked again same ans whats wrong? :confused:

    its 2/cos2theta ?
    Basically,

    \dfrac{d}{d\theta}tan(\theta) \neq sec(\theta)

    which is what you put in the first post. It's what you've calculated for the very first part. I suspect this mistake was carried through into your calculation of \dfrac{d}{d\theta}tan(2\theta) which is why you're getting the wrong answer.
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    (Original post by EEngWillow)
    Basically,



    which is what you put in the first post. It's what you've calculated for the very first part. I suspect this mistake was carried through into your calculation of which is why you're getting the wrong answer.
    I'm sorry I'm completely LOST :confused: can you please post The working for some bits at least, since i dont want to take any more of your time
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    u can....
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    Can somebody atleast give me the solution EVEN without no working, so that i could work towards it
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    (Original post by mufas)
    its tan 2 theta not just tan theta
    yeah, you can use the chain rule afterwards.

    answer you want is 2sec²(2x)

    Spoiler:
    Show

    il walk you through it.
    you want to differentiate u/v

    you have u=sin(2x) and v=cos(2x)

    quotient rule says the derivative is: (u'.v - u.v')/v²

    which is: [ 2cos(2x).cos(2x) - -2sin(2x).sin(2x) ] / cos²(2x) = [ 2cos²(2x) + 2sin²(2x) ]/cos²(2x) = 2[ cos²(2x) + sin²(2x) ]/cos²(2x) = 2/cos²(2x) = 2sec²(2x)
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    (Original post by RK92)
    yeah, you can use the chain rule afterwards.

    answer you want is 2sec²(2x)

    Spoiler:
    Show

    il walk you through it.
    you want to differentiate u/v

    you have u=sin(2x) and v=cos(2x)

    quotient rule says the derivative is: (u'.v - u.v')/v²

    which is: [ 2cos(2x).cos(2x) - -2sin(2x).sin(2x) ] / cos²(2x) = [ 2cos²(2x) + 2sin²(2x) ]/cos²(2x) = 2[ cos²(2x) + sin²(2x) ]/cos²(2x) = 2/cos²(2x) = 2sec²(2x)
    I have got this same answer earlier on but it doesn't work for some reason

    (1/cosx)/(2sec²(2x) ) = 2cosx/ cos2x so when you sub the pi/6 you dont get 6
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    (Original post by mufas)
    I have got this same answer earlier on but it doesn't work for some reason

    (1/cosx)/(2sec²(2x) ) = 2cosx/ cos2x so when you sub the pi/6 you dont get 6
    I'll repeat what I said then. In your first post, you've said this:

    I differentiated tan theta got 1/cos theta
    This is not correct. You need to fix this before you can get the right answer.
 
 
 
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