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    Attachment 610232610234 hi, I am wondering whether I have differentiated this correctly or not . I couldn't check this expect from subbing the x and y values in .. Is there other ways to check it on the calculator ?

    Thanks
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    (Original post by coconut64)
    Attachment 610232610234 hi, I am wondering whether I have differentiated this correctly or not . I couldn't check this expect from subbing the x and y values in .. Is there other ways to check it on the calculator ?

    Thanks
    When using implicit differentiation, you differentiate with respect to x the whole expression so, as you used the Quotient Rule, your u= sin x and v= sin y. Now, when you differentiate u, it will be cos x but as you differentiate v, it will be cos y dy/dx because you're differentiating with respect to x. Whenever you are differentiating a y term, you must always write dy/dx after differentiating else you lose marks.
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    (Original post by coconut64)
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    This is correct.
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    (Original post by sabahshahed294)
    This is correct.
    The differentiation looks good but the denominator of the gradient isn't 5 when you sub in x = 0.
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    (Original post by BrasenoseAdm)
    The differentiation looks good but the denominator of the gradient isn't 5 when you sub in x = 0.
    Oo yeah, I read that as y by mistake. My bad.
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    (Original post by BrasenoseAdm)
    The differentiation looks good but the denominator of the gradient isn't 5 when you sub in x = 0.
    So the gradient is just 2? Thanks
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    (Original post by coconut64)
    hi, I am wondering whether I have differentiated this correctly or not . I couldn't check this expect from subbing the x and y values in .. Is there other ways to check it on the calculator ?
    With the \dfrac{\sin x}{\sin y} = 2 question, rather than use the quotient rule, it's *much* easier to go:

    \dfrac{\sin x}{\sin y} = 2 \implies \sin x = 2 \sin y

    and then differentiate the 2nd equation implicitly instead.
 
 
 
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