Question

Find dy/dx by implicit differentiation. (sin(pi*x) + cos(pi*y))^2= 2. So far I am up to 2pi (sin(pi*x)+ cos(pi*y)) * (cos(pi*x) - dy/dx sin(pi*y) = 0. What do I do from here?

Answer 1

\[\Large\rm \sin(\pi x)+\cos^2(\pi y)=2\]Differentiating gives us,\[\Large\rm \cos(\pi x)\color{royalblue}{(\pi x)'}+2 \cos(\pi y)\color{royalblue}{\left[\cos(\pi y)\right]'}=0\]So that's step, the blue parts we need to differentiate now, due to the chain rule.\[\Large\rm \pi \cos(\pi x)+2 \cos(\pi y)\left[-\sin(\pi y)\right]\color{royalblue}{(\pi y)'}=0\]I'm sure you're past this point, I just need to catch up a sec :)

Answer 2

\[\Large\rm \pi \cos(\pi x)+2 \cos(\pi y)\left[-\sin(\pi y)\right](\pi y')=0\]\[\Large\rm \pi \cos(\pi x)-2 \pi y' \sin(\pi y) \cos(\pi y)=0\]

Answer 3

Divide each side by pi,\[\Large\rm \cos(\pi x)-2y' \sin(\pi y) \cos(\pi y)=0\]Add that big ... second term.. to the other side,\[\Large\rm \cos(\pi x)=2y' \sin(\pi y)\cos(\pi y)\]

Answer 4

Oh the square was on the whole thing? .... :(

Answer 5

Awesome...

Answer 6

So hard to read the stupid text :( UGHHHHH

Answer 7

Wish you had stopped me earlier :c

Answer 8

Yes sorry I was trying to catch you but you were moving fast.

Answer 9

\[\Large\rm \left(\sin(\pi x) + \cos(\pi y)\right)^2= 2\]Differentiating:\[\Large\rm 2\left[\sin(\pi x) + \cos(\pi y)\right]\color{royalblue}{\left[\sin(\pi x) + \cos(\pi y)\right]'}=0\]Chain rule,\[\Large\rm 2\left[\sin(\pi x) + \cos(\pi y)\right]\color{orangered}{\left[\pi\cos(\pi x) -\pi y'\sin(\pi y)\right]}=0\]Mmm ok, that look better? :o

Answer 10

And it looks like you factored a pi out from that point.
Ok soooo

Answer 11

Yes. That looks better. This section totally lost me so I'm not sure if I was suppose to factor the pi out.

Answer 12

We could divide each side by 2pi,\[\Large\rm (\sin \pi x+\cos \pi y)(\cos \pi x- y' \sin \pi y)=0\]I wanna sayyyyyyy that we should just divide each side by (sin pix + cos piy), but I'm worried we'll lose some information if we do that.
We would have to assume that sin pix+cos pi y can not be zero.

So let's just avoid doing that for now, expand out the brackets.

Answer 13

\[\rm \sin \pi x \cos \pi x + \cos \pi x \cos \pi y- y' \sin \pi y \cos \pi y- y' \sin \pi x \sin \pi y =0\]So we get something like that, yes? >.<

Answer 14

Factoring a y' out of each of those last terms:\[\rm \sin \pi x \cos \pi x + \cos \pi x \cos \pi y- y'(\sin \pi y \cos \pi y- \sin \pi x \sin \pi y) =0\]

Answer 15

Then solve for y'
! :)

Answer 16

Subtract some stuff, divide some stuff...
What'd you think? Too scary?

Answer 17

Okay. Awesome. I can take it from here. Thank you so much!

Answer 18

cool c: