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How to differentiate x = cos(2 y + pi ) ?? help please!! core 3!! Watch

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    Please can someone help me with the above question, i am really struggling!
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    differentiate with respect to y, so you get dx/dy = ...

    then reciprocate so you are left with dy/dx = ...
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    (Original post by stanmoor)
    Please can someone help me with the above question, i am really struggling!
    With respect to x or y?
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    are you differentiating with respect to y?
    just use the chain rule
    x=cos(2y +pi)
    x=cos(u), u=2y +pi
    dx/dy=dx/du * du/dy
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    i have to find dy/dx
    why do you put x = cos(u) ??
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    Alternatively,

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     arccos(x) = 2y+ \pi

     y= \dfrac{1}{2} arccos(x) - \dfrac{1}{2} \pi

     \dfrac{dy}{dx} = \dfrac{-1}{2\sqrt{1-x^2}}
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    (Original post by stanmoor)
    Please can someone help me with the above question, i am really struggling!
    Are you differentiating both sides with respect to x or with respect to y? There are two variables in the equation so you could differentiate with respect to either of them.

    If you differentiate with respect to y then the left-hand side is simply dx/dy. The right-hand side requires use of the chain-rule. Substitute another variable, say u, in place of (2y+pi). You should be able to differentiate the cosine function of this variable (u) with respect to u as well as u with respect to x. This is all the chain-rule requires.

    If you wish to differentiate both sides with respect to x then implicit differentiation can be used.

    I hope this has helped. If not, please quote me back.
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    so can you use the product rule
    using u as cos and v as 2y + pi??

    you can't just have cos on its on though can you??
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    If we differentiate cos(f(x)) then we gets -(f'(x))(sinf(x))

    also know that 1/dx/dy = dy/dx

    We don't need to know how to derive this from the chain rule in c3, we just need to know the formula. You won't be getting any extra marks, but it's pretty easy to derive the formula, so if you want learn the formula
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    chain rule
 
 
 
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