# Implicit differentiation

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#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
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3 years ago
#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.
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3 years ago
#4
(Original post by coconut64)
Also this one
This is correct.
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3 years ago
#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.
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3 years ago
#6
The differentiation looks good but the denominator of the gradient isn't 5 when you sub in x = 0.
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#7
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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3 years ago
#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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