Question

Find y' if arctan(xy)=8+x^2y

Answer 1

Hmm ok to start, do you recall the derivative of arctan(x)? :)

Answer 2

1/1+x^2

Answer 3

Ok good good good.
\[\Large \left[\arctan(\color{royalblue}{x})\right]'\quad=\quad \frac{1}{1+(\color{royalblue}{x})^2}\]So the derivative of arctan(xy) should give us:\[\Large \left[\arctan(\color{royalblue}{xy})\right]'\quad=\quad \frac{1}{1+(\color{royalblue}{xy})^2}\]But since our argument is more than just x, we `need` to apply the chain rule.\[\Large \left[\arctan(\color{royalblue}{xy})\right]'\quad=\quad \frac{1}{1+(\color{royalblue}{xy})^2}(xy)'\]
Multiplying by the derivative of the inner function.

Answer 4

So for our inner function here, we need to take it's derivative.
Looks like we'll want to apply the product rule, yes?

Answer 5

Yea, I understood that and then take the derivative of the other side. But it gets really long, so I feel like I just missed an x or y....

Answer 6

Ok left side gives us:\[\Large \frac{1}{1+(xy)^2}(y+xy')\]Look ok so far? :o

Answer 7

And the right side,
\[\Large (8+x^2y)'\quad=\quad 2xy+x^2y'\]

Answer 8

Yea

Answer 9

Ya solving for y' is a bit of a pain from there I guess :)
Hmm let's see.

One thing that would make this problem a bit easier is if we multiply through by that denominator.\[\Large \frac{1}{1+(xy)^2}(y+xy')\quad=\quad 2xy+x^2y'\]Giving us:\[\Large y+xy'\quad=\quad 2xy\left[1+(xy)^2\right]+x^2\left[1+(xy)^2\right]y'\]

Answer 10

Get all of the `y'` to one side, other stuff on the other side,\[\Large xy'-x^2\left[1+(xy)^2\right]y'\quad=\quad 2xy\left[1+(xy)^2\right]-y\]

Answer 11

From there we would just factor a y' out of each term on the left, and divide by the leftover. Do you have an answer key or something to compare? :D
Hopefully this is closer to what we're looking for.

Answer 12

Yes, thank you very much!

Answer 13

cool