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    Given that x= 4sin(2y+6), find dy/dx in terms of x.
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    Easy first you need to differentiate dx/dy NOT dy/dx
    then knowing the fact that dy/dx=1/dx/dy then that is your answer:P

    x=4sin(2y+6)

    dx/dy=4x2cos(2y+6) remember sinf(x)=f'(x)cosf(x)

    dx/dy=8cos(2y+6) >> dy/dx=1/8cos(2y+6)
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    (Original post by Remarqable M)
    Easy first you need to differentiate dx/dy NOT dy/dx
    then knowing the fact that dy/dx=1/dx/dy then that is your answer:P
    But it wants the answer in terms of x, the equation is given in terms of y.
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    (Original post by WhatIsAUsername?)
    But it wants the answer in terms of x, the equation is given in terms of y.
    Re-arrange the equation to make it in terms of x, and substitute it back into the differential.
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    (Original post by Mathematician!)
    Re-arrange the equation to make it in terms of x, and substitute it back into the differential.
    is my answer correct? did i do something wrong?
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    (Original post by Remarqable M)
    is my answer correct? did i do something wrong?
    Your answer is correct, but it's incomplete (as the OP has pointed out).
    Just follow my above advice to get to the final answer.
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    (Original post by Mathematician!)
    Your answer is correct, but it's incomplete (as the OP has pointed out).
    Just follow my above advice to get to the final answer.
    could you do this question both for me and OP please:p: in latex form
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    ORIGINAL QUESTION: x= 4sin(2y+6), find dy/dx in terms of x.

    ANSWER:
    Spoiler:
    Show
     \displaystyle\frac{dx}{dy} = 8 \cos (2y+6)

    \displaystyle \frac{dy}{dx} = \frac{1}{8 \cos (2y+6)}

    And as  \displaystyle y = \frac{1}{2} \sin^{-1}(\frac{x}{4}) - 3 ,

    \displaystyle \frac{dy}{dx} = \frac{1}{8\cos(\sin^{-1}(\frac{x}{4}))}
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    (Original post by Mathematician!)
    ORIGINAL QUESTION: x= 4sin(2y+6), find dy/dx in terms of x.

    ANSWER:
    Spoiler:
    Show
     \displaystyle\frac{dx}{dy} = 8 \cos (2y+6)

    \displaystyle \frac{dy}{dx} = \frac{1}{8 \cos (2y+6)}

    And as  \displaystyle y = \frac{1}{2} \sin^{-1}(\frac{x}{4}) - 3 ,

    \displaystyle \frac{dy}{dx} = \frac{1}{8\cos(\sin^{-1}(\frac{x}{4}))}
    The answer should not include cos or the inverse of sin. What would the answer be without them.
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    (Original post by WhatIsAUsername?)
    The answer should not include cos or the inverse of sin. What would the answer be without them.
    Start with  \displaystyle y = \frac{1}{2} \sin^{-1}(\frac{x}{4}) - 3

    Now \displaystyle \frac{dy}{dx} = 0.5\frac{d (\sin^{-1}(\frac{x}{4}))}{dx} + \frac{d(-3)}{dx}

    So, \displaystyle 2\frac{dy}{dx} = \frac{d (\sin^{-1}(\frac{x}{4}))}{dx}

    Use the chain rule here, by substituting \displaystyle u = \frac{x}{4}.

    So \displaystyle 2\frac{dy}{dx} =\frac{d (\sin^{-1}u)}{du}\times\frac{d(\frac{x}{  4})}{dx}

    These utilise simple differentiation, and the general rule for the differential of inverse sine functions (though it can be derived using trigonometric identities. Ask me if you want me to show this).

    I'll leave the rest to you.
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    (Original post by Mathematician!)
    Start with  \displaystyle y = \frac{1}{2} \sin^{-1}(\frac{x}{4}) - 3

    Now \displaystyle \frac{dy}{dx} = 0.5\frac{d (\sin^{-1}(\frac{x}{4}))}{dx} + \frac{d(-3)}{dx}

    So, \displaystyle 2\frac{dy}{dx} = \frac{d (\sin^{-1}(\frac{x}{4}))}{dx}

    Use the chain rule here, by substituting \displaystyle u = \frac{x}{4}.

    So \displaystyle 2\frac{dy}{dx} =\frac{d (\sin^{-1}u)}{du}\times\frac{d(\frac{x}{  4})}{dx}

    These utilise simple differentiation, and the general rule for the differential of inverse sine functions (though it can be derived using trigonometric identities. Ask me if you want me to show this).

    I'll leave the rest to you.
    Although not covered in C3(the differential of inverse sine functions), I appreciate your working and the effort you've put into explaining this simple differentiation, which i will need in the future ofcourse I will try this alternative method out it looks simple. Thanks again
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    (Original post by Remarqable M)
    Although not covered in C3(the differential of inverse sine functions), I appreciate your working and the effort you've put into explaining this simple differentiation, which i will need in the future ofcourse I will try this alternative method out it looks simple. Thanks again
    Well I can show you how it works! It is rather simple (when you know how! :p:) Open the following spoiler if you would like to see how it's done...

    Spoiler:
    Show
    Let  y = \sin^{-1} x

    It follows, of course, that  x = \sin y

    So using either implicit differentiation, or direct differentiation, you get  \displaystyle \frac{dy}{dx} = \frac{1}{\cos y}

    Now this can be reduced further, as we know that  \sin^2 y + \cos^2 y = 1

    So this re-arranges to give  \cos y = \sqrt{1 - sin^2 y}

    Now using x = \sin y , it follows that  \cos y = \sqrt{1 - x^2}

    This can be subsituted into the differential to give  \displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}
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    (Original post by Mathematician!)
    Well I can show you how it works! It is rather simple (when you know how! :p:) Open the following spoiler if you would like to see how it's done...

    Spoiler:
    Show
    Let  y = \sin^{-1} x

    It follows, of course, that  x = \sin y

    So using either implicit differentiation, or direct differentiation, you get  \displaystyle \frac{dy}{dx} = \frac{1}{\cos y}

    Now this can be reduced further, as we know that  \sin^2 y + \cos^2 y = 1

    So this re-arranges to give  \cos y = \sqrt{1 - sin^2 y}

    Now using x = \sin y , it follows that  \cos y = \sqrt{1 - x^2}

    This can be subsituted into the differential to give  \displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}
    :woo: thanks it makes sense now, how the rule is derived.
 
 
 
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