Question

Find dy/dx by implicit differentiation.
cotxy+xy=0

Answer 1

Ooo this is a cool problem :)
Let's see what matty mat mat boy comes up with

Answer 2

\[\cot(xy)+xy=0\]\[-\csc ^2(xy)(xy'+y)+xy'+y=0\]\[-xy' \csc^2(xy)+y \csc^2(xy)+xy'+y=0\]\[-xy' \csc^2 (xy)+xy'=-y \csc^2(xy)-y\]\[-xy' [\csc^2(xy)-1]=y [\csc^2(xy)-1]\]\[y'=-{y \over x}\]

Answer 3

How did matty mat mat boy do? :P

Answer 4

Looks good >.<
Where's some explanation though bruhhh? lol

Answer 5

Haha I know, just wanted to lay down some calc...

STEP #1: Apply chain rule (with product rule) to cot(xy). Apply product rule to xy. 0 stays the same on the right side.

STEP #2: We want to isolate the y' (same as dy/dx). To do that, we need to collect all y' terms on the same side. To do THAT, we need to distribute -csc^2(xy).

STEP #3: Group y' terms on the left. All terms that don't have a y' are transposed.

STEP #4 and #5: Factor out the GCF on both sides. I noticed that factoring out more than just y' actually makes the equation much easier. If you factor out a -x on the left in addition to y', and you factor out y on the right, you'll notice that what's inside the square brackets is the same on both sides. You can remove this from the equation entirely. Then just divide both sides by -x to isolate y'.

STEP #6: Breathe.

Answer 6

Thanks for the explanation! @matt101