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Implicit differentiation watch

1. 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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2. Also this one
3. (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.
4. (Original post by coconut64)
Also this one
This is correct.
5. (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.
6. (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.
7. (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
8. (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 question, rather than use the quotient rule, it's *much* easier to go:

and then differentiate the 2nd equation implicitly instead.

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