find dy/dx of sin(xy)=x+y
So far I have
cos(xy)*(xy'+1y)=1+y'
(cos(xy))xy'+(cos(xy))y=1+y'
Then, I'm sure you solve here for y but for the life of me I don't see the logical steps to get to y.
Would it be
y=((1+y')-(cos(xy)xy'))/(cos(xy))
???
find dy/dx of sin(xy)=x+y
So far I have
cos(xy)*(xy'+1y)=1+y'
(cos(xy))xy'+(cos(xy))y=1+y'
Then, I'm sure you solve here for y but for the life of me I don't see the logical steps to get to y.
Would it be
y=((1+y')-(cos(xy)xy'))/(cos(xy))
???
$\displaystyle \dfrac{d}{dx}\left[\sin(xy) = x+y]$
$\displaystyle \cos(xy) \cdot (xy' + y) = 1 + y'$
distribute the $\displaystyle \cos(xy)$ on the left ...
$\displaystyle xy'\cos(xy) + y\cos(xy) = 1 + y'$
get the terms w/ $\displaystyle y'$ on the same side ...
$\displaystyle xy'\cos(xy) - y' = 1 - y\cos(xy)$
factor out $\displaystyle y'$ ...
$\displaystyle y'[x\cos(xy) - 1] = 1 - y\cos(xy)$
$\displaystyle y' = \dfrac{1 - y\cos(xy)}{x\cos(xy) - 1}$