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    Ok so the first part of the question is x = 3sin y find DX/DY in terms of Y
    I got 3cos y which is right, then it asked for me to find DY/DX in terms of Y, i then got 1/3 Sec Y, the next part is hence show DY/DX = 1/Root 9 - x to the power of 2, I cant do this part I dont know how to get from 1/3 Sec y to that?
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    (Original post by SunDun111)
    Ok so the first part of the question is x = 3sin y find DX/DY in terms of Y
    I got 3cos y which is right, then it asked for me to find DY/DX in terms of Y, i then got 1/3 Sec Y, the next part is hence show DY/DX = 1/Root 9 - x to the power of 2, I cant do this part I dont know how to get from 1/3 Sec y to that?
    Hint: \cos^2(y)=1-\sin^2(y).
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    (Original post by joostan)
    Hint: \cos^2(y)=1-\sin^2(y).
    Ey but, isn't it asking me to get from 1/3 Sec y to 1/ Root 9 - x to the power of 2 ,
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    (Original post by SunDun111)
    Ey but, isn't it asking me to get from 1/3 Sec y to 1/ Root 9 - x to the power of 2 ,
    Yes, but \sec(y)=\dfrac{1}{\cos(y)}. . .
    Also worth noting that a\sqrt{f(x)}=\sqrt{a^2f(x)}.
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    (Original post by joostan)
    Yes, but \sec(y)=\dfrac{1}{\cos(y)}. . .
    I understand, but I'm asking what is the question asking of me? Is it asking me to rearrrange DY/DX in terms of Y to in terms of X?
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    (Original post by SunDun111)
    ...
    The question wants you to go from \frac{dy}{dx} = \frac{1}{3\cos y} into dy/dx as a function of x using the fact that x = 3\sin y
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    (Original post by Zacken)
    The question wants you to go from \frac{dy}{dx} = \frac{1}{3\sec y} into dy/dx as a function of x using the fact that x = 3\sin y
    Right thanks!
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    (Original post by SunDun111)
    Right thanks!
    Do you now understand Joostan's (excellent) advice and can apply it here?
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    (Original post by Zacken)
    Do you know understand Joostan's (excellent) advice and can apply it here?
    First time i've heard it but trying to,
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    (Original post by SunDun111)
    First time i've heard it but trying to,
    Got it?
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    (Original post by Zacken)
    Got it?
    Unforunatley not, i understand the concept and have done other questions like this just now, but this particular one I am struggling to re-arrange, I will ask my teacher when I go back on Monday
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    (Original post by SunDun111)
    Unforunatley not, i understand the concept and have done other questions like this just now, but this particular one I am struggling to re-arrange, I will ask my teacher when I go back on Monday
    \displaystyle 

\begin{equation*}\frac{1}{3\cos y} = \frac{1}{3\sqrt{1 - \sin^2 y}} \end{equation*}

    since \cos^2 y + \sin^2 y = 1 \Rightarrow \cos^2 y = 1 - \sin^2 y \Rightarrow \cos y = \sqrt{1 - \sin^2 y}

    Now you know that x = 3\sin y so \frac{x}{3} = \sin y and \sin^2 y = \frac{x^2}{9}.

    Plug this back into the centered equation and remember that 3\sqrt{1 -\sin^2 y} \equiv \sqrt{9 - 9\sin^2 y}.
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    (Original post by Zacken)
    \displaystyle 

\begin{equation*}\frac{1}{3\cos y} = \frac{1}{3\sqrt{1 - \sin^2 y}} \end{equation*}

    since \cos^2 y + \sin^2 y = 1 \Rightarrow \cos^2 y = 1 - \sin^2 y \Rightarrow \cos y = \sqrt{1 - \sin^2 y}

    Now you know that x = 3\sin y so \frac{x}{3} = \sin y and \sin^2 y = \frac{x^2}{9}.

    Plug this back into the centered equation and remember that 3\sqrt{1 -\sin^2 y} \equiv \sqrt{9 - 9\sin^2 y}.
    Thanks I got it right, quick question
    i am solving the equation 1 - 2ln X I took the 2lnx to the other side, and E'd both sides so i got
    E to the power of 1 = 2x
    Is the answer e/2?
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    (Original post by SunDun111)
    Thanks I got it right, quick question
    i am solving the equation 1 - 2ln X I took the 2lnx to the other side, and E'd both sides so i got
    E to the power of 1 = 2x
    Is the answer e/2?
    No you didn't.

    \displaystyle 

\begin{equation*}1 - 2\ln x = 0 \Rightarrow 1=  2\ln x \Rightarrow e^1 = e^{2\ln x} \Rightarrow e = e^{\ln x^2} \Rightarrow x^2 = e \Rightarrow \cdots \end{equation*}

    It is not the case that e^{a \ln f(x)} = a f(x).

    It is the case that e^{a\ln f(x)} = e^{\ln f(x)^a} = f(x)^a.
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    (Original post by Zacken)
    No you didn't.

    \displaystyle 

\begin{equation*}1 - 2\ln x = 0 \Rightarrow 1=  2\ln x \Rightarrow e^1 = e^{2\ln x} \Rightarrow e = e^{\ln x^2} \Rightarrow x^2 = e \Rightarrow \cdots \end{equation*}

    It is not the case that e^{a \ln f(x)} = a f(x) it is the case that e^{a\ln f(x)} = e^{\ln f(x)^a} = f(x)^a.
    I see where I've gone wrong thanks
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    (Original post by SunDun111)
    I see where I've gone wrong thanks
    No problem.
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    (Original post by Zacken)
    No problem.
    Sorry for all the questions, doing a lot of revision today

    This question is the curve has the equation (x to the power of 2 - 4x +1)e to the minus x, Dy/DX it and find the points,

    Dy/dx should be minus e to the minus x (x to the power of 2 - 4x +1) + e to the minus x (2x - 2) = 0 how would I solve this?
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    (Original post by SunDun111)
    Sorry for all the questions, doing a lot of revision today

    This question is the curve has the equation (x to the power of 2 - 4x +1)e to the minus x, Dy/DX it and find the points,

    Dy/dx should be minus e to the minus x (x to the power of 2 - 4x +1) + e to the minus x (2x - 2) = 0 how would I solve this?
    \displaystyle 

\begin{equation*}e^{-x}(x^2 - 4x + 1) + e^{-x}(2x - 2) = e^{-x}(x^2 - 4x + 2x - 2) = 0\end{equation*}

    So e^{-x} = 0 (no solutions) or x^2 - 4x + 2x - 2 = 0 (two solutions).
 
 
 
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