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    If x=cosy, find dy/dx in terms of x. I can do it in terms of y but I'm not sure how to get it in terms of x?

    The answer is -1/(1-x^2)^0.5
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    (Original post by Daisy127)
    If x=cosy, find dy/dx in terms of x. I can do it in terms of y but I'm not sure how to get it in terms of x?

    The answer is -1/(1-x^2)^0.5
    few clues:

    - implicit differntitation
    -(sinx)^2+(cosx)^2=1
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    @imran___
    Can we not say

    If X=COS(y)

    Then doing the differential of x in respect to y we would get
    dx/dy = -SIN(y)

    Doing the insverse of both sides
    dy/dx = 1/ [-SIN(y)]
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    (Original post by Daisy127)
    If x=cosy, find dy/dx in terms of x. I can do it in terms of y but I'm not sure how to get it in terms of x?

    The answer is -1/(1-x^2)^0.5
    Use cos^2y +sin^2y=1. in place of cos^2y, you can put x^2 as x=cosy and rearrange for siny. to find siny in terms of x
    you know dy/dx = -1/siny , if you have done it correctly, you can substitute -siny in terms of x.
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    (Original post by Daisy127)
    If x=cosy, find dy/dx in terms of x. I can do it in terms of y but I'm not sure how to get it in terms of x?

    The answer is -1/(1-x^2)^0.5
    I don't know why you were presented with the equation as x=cos(y). However, if you rearrange the equation to y=cos^-1(x), you can then differentiate with respect to x.

    If you're doing AQA, the differential of the inverse cos of x is provided in the formula book.

    Therefore, dy/dx of cos^-1(x) is -1/root(1-x)
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    (Original post by Daisy127)
    If x=cosy, find dy/dx in terms of x. I can do it in terms of y but I'm not sure how to get it in terms of x?The answer is -1/(1-x^2)^0.5
    differentiate the function x with respect to y to get dx/dy then 'flip' it
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    (Original post by User1212)
    differentiate the function x with respect to y to get dx/dy then 'flip' it
    Thats what I thought but you would get 1/[1-sin(2y)]
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    Not sure if my first post was clear

     cos^2y +sin^2y =1

     sin^2y=1-cos^2y

     siny = (1 - cos^2y)^{\frac{1}{2}} then  x= cosy , x^2= cos^2y then you can find dy/dx in terms of x.
    and clearly;

     siny = (1-x^2)^{\frac{1}{2}}
    Or you can use triangles.
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    (Original post by c1s2)
    Thats what I thought but you would get 1/[1-sin(2y)]
    yeah I see what you mean, I should have read the question properly lol
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    (Original post by c1s2)
    Thats what I thought but you would get 1/[1-sin(2y)]
    No, x=cosy, so \frac{dx}{dy} = -siny now "flip it" :  \frac{dy}{dx} = \frac{1}{-siny}
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    (Original post by NotNotBatman)
    No, x=cosy, so \frac{dx}{dy} = -siny now "flip it" :  \frac{dy}{dx} = \frac{1}{-siny}

    My error, I miss read the question to be x=cos(2y)
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    (Original post by c1s2)
    My error, I miss read the question to be x=cos(2y)
    Do you know how to proceed?
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    (Original post by NotNotBatman)
    No, x=cosy, so \frac{dx}{dy} = -siny now "flip it" :  \frac{dy}{dx} = \frac{1}{-siny}
    Hey, thanks for the help. There are more questions like this further on in the book that say to use this inverse method...how would you finish the question in this way?
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    (Original post by Daisy127)
    Hey, thanks for the help. There are more questions like this further on in the book that say to use this inverse method...how would you finish the question in this way?
    In C3? What exam board?

    Not entirely sure what that would mean, if you had y = arccosx, you would either use the standard derivative in a formula book or take cosine to both sides and then you would use implicit differentiation.

     x = cos y, diff wrt x , 1= -siny \frac{dy}{dx},

    \frac{dy}{dx} =-\frac{1}{siny} = -\frac{1}{\sqrt{1-x^2}}
 
 
 
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