Question

Check out my answer find dy/dx using implicit differentiation xe^y = ysinx
I got e^y-ycos(x)/sin(x)-xe^y

Answer 1

\[\Large x e^y\quad=\quad y \sin x\]Taking the derivative of both sides, with respect to x,\[\Large e^y+xe^y y'\quad=\quad y' \sin x + y \cos x\]Then we do some stuff...\[\Large y'(x e^y-\sin x)=y \cos x-e^y\]Solving for y',\[\Large y'=\frac{y \cos x - e^y}{xe^y-\sin x}\]

Answer 2

If I were to factor a negative out of the numerator AND denominator, we would get,\[\Large y'\quad=\quad \frac{e^y-y \cos x}{\sin x-xe^y}\]

Answer 3

Yup, looks like you've got everything figured out! :)

Answer 4

Ooo ooo careful :x keep that to yourself!
Don't use the word test around here, that's cheatin broski!

Answer 5

Ah didn't know thanks for the tip