C3 Implicit Differentiation Help Watch

starwarsjedi123
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#1
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Hi.
I am doing a C3 paper (link: http://www.mei.org.uk/files/papers/c308ju_fg48.pdf)

Anyway, i am stuck on Q7. I am ok with the method of implicit differentiation. Differentiate x terms and y terms, and stick a dy/dx after the y terms. Then you rearrange etc. However, i am slightly confused when on the part where i need to differentiate xy. I would use the product rule, however, when using the product rule, im unsure where the dy/dx goes

so i have x^2 + xy+y^2=12
i would do 2x+x+y(dy/dx)+2y(dy/dx)=0
but this is wrong
Help??
thanks
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Kvothe the Arcane
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You have differentiated xy wrtx wrongly I'm afraid.

\dfrac{\mathrm{d}}{\mathrm{d}x}x  y=x \times \dfrac{\mathrm{d}}{\mathrm{d}x}y + y \times \dfrac{\mathrm{d}}{\mathrm{d}x}x

Do you see why that's the case?

Can you do the next step?

Also, moved to maths.
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Serine Soul
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(Original post by starwarsjedi123)
Hi.
I am doing a C3 paper (link: http://www.mei.org.uk/files/papers/c308ju_fg48.pdf)

Anyway, i am stuck on Q7. I am ok with the method of implicit differentiation. Differentiate x terms and y terms, and stick a dy/dx after the y terms. Then you rearrange etc. However, i am slightly confused when on the part where i need to differentiate xy. I would use the product rule, however, when using the product rule, im unsure where the dy/dx goes

so i have x^2 + xy+y^2=12
i would do 2x+x+y(dy/dx)+2y(dy/dx)=0
but this is wrong
Help??
thanks
Look at this bit again

Kvothe's explained for me :awesome:
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the bear
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xy =======> x*dy/dx + y*1
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starwarsjedi123
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(Original post by Serine Soul)
Look at this bit again

aaaah ok i understand. i didnt really fully understand 'why' i was doing what i was. Thanks kvoothe
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Kvothe the Arcane
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(Original post by starwarsjedi123)
Ok. i have differentiated 'xy' using the product rule, which gives me x+y.
Surely i stick the dy/dx after the y???
Have you looked at my post?

You don't just stick dy/dx infront of a variable.

(Original post by Kvothe the arcane)

\dfrac{\mathrm{d}}{\mathrm{d}x}x  y=x \times \dfrac{\mathrm{d}}{\mathrm{d}x}y + y \times \dfrac{\mathrm{d}}{\mathrm{d}x}x

Do you see why that's the case?

Can you do the next step?
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Serine Soul
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(Original post by starwarsjedi123)
Ok. i have differentiated 'xy' using the product rule, which gives me x+y.
Surely i stick the dy/dx after the y???
So with implicit differentiation, when you used the product rule, dy/dx of y alone will simply be dy/dx (1)
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Zacken
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(Original post by starwarsjedi123)
Ok. i have differentiated 'xy' using the product rule, which gives me x+y.
Surely i stick the dy/dx after the y???
Do you know how the product rule works? \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(x  y) = \frac{\mathrm{d}}{\mathrm{d}x}(x  )y + x\frac{\mathrm{d}}{\mathrm{d}x}(  y).

We also have that \frac{\mathrm{d}}{\mathrm{d}x}(y  ) = \frac{\mathrm{d}y}{\mathrm{d}x}.
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Zacken
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(Original post by Serine Soul)
So with implicit differentiation, when you used the product rule, dy/dx of y alone will simply be dy/dx (1)
Might want to be careful there, it's \frac{\mathrm{d}}{\mathrm{d}x} of y - the derivative is an operator.
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Serine Soul
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(Original post by Zacken)
Might want to be careful there, it's \frac{\mathrm{d}}{\mathrm{d}x} of y - the derivative is an operator.
Lol alright

I can't explain math online
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