The Student Room Group

C3 Implicit Differentiation Help

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
(edited 8 years ago)
You have differentiated xy wrtx wrongly I'm afraid.

ddxxy=x×ddxy+y×ddxx\dfrac{\mathrm{d}}{\mathrm{d}x}xy=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.
(edited 8 years ago)
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:
xy =======> x*dy/dx + y*1
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
(edited 8 years ago)
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? :h:

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

Original post by Kvothe the arcane


ddxxy=x×ddxy+y×ddxx\dfrac{\mathrm{d}}{\mathrm{d}x}xy=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?
(edited 8 years ago)
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)
Reply 7
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? ddx(xy)=ddx(x)y+xddx(y)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(xy) = \frac{\mathrm{d}}{\mathrm{d}x}(x)y + x\frac{\mathrm{d}}{\mathrm{d}x}(y).

We also have that ddx(y)=dydx\frac{\mathrm{d}}{\mathrm{d}x}(y) = \frac{\mathrm{d}y}{\mathrm{d}x}.
Reply 8
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 ddx\frac{\mathrm{d}}{\mathrm{d}x} of yy - the derivative is an operator.
Original post by Zacken
Might want to be careful there, it's ddx\frac{\mathrm{d}}{\mathrm{d}x} of yy - the derivative is an operator.


Lol alright

I can't explain math online

Quick Reply

Latest