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Implicit differentiation

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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
14840462868781770128893.jpg343.9KB
1484046312830-738677661.jpg346.9KB
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coconut64
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Also this one 1484046489493983524854.jpg
Original post by coconut64
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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.
(edited 7 years ago)
Original post by coconut64
Also this one 1484046489493983524854.jpg


This is correct. :smile:
Original post by sabahshahed294
This is correct. :smile:


The differentiation looks good but the denominator of the gradient isn't 5 when you sub in x = 0.
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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coconut64
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16
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
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 sin⁡xsin⁡y=2\dfrac{\sin x}{\sin y} = 2 question, rather than use the quotient rule, it's *much* easier to go:

sin⁡xsin⁡y=2  ⟹  sin⁡x=2sin⁡y\dfrac{\sin x}{\sin y} = 2 \implies \sin x = 2 \sin y

and then differentiate the 2nd equation implicitly instead.

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