Question

Find dy/dx if xe^y+1=xy

Answer 1

\[\Large y(x) \cdot x = 1 + xe^{y(x)} \]
Writing down the arguments so you see that you have to apply the chain rule.

Answer 2

Keeping that in mind, you will end up with something like :

\[\Large y'x+y=e^y+e^yy'x \]

Answer 3

\[x e^{y} + 1 = x y\]take derivative of both sides with respect to x :
\[\frac{ d }{ dx }(x e^{y} + 1 )=\frac{ d }{dx }(xy)\]sum of derivatives is the derivative of the sum :\[\frac{ d }{ dx }(x e^{y}) + \frac{d}{dx}(1 )=\frac{ d }{dx }(xy)\]derviative of a constant is zero :\[\frac{ d }{ dx }(x e^{y}) + 0 =\frac{ d }{dx }(xy)\]product rule :\[e^{y}\frac{ d }{ dx }(x) + x\frac{d}{dx} (e^{y}) =y\frac{ d }{dx }(x)+ x\frac{ d }{dx }(y)\]simplify with derivative of x with respect to itself is one :\[e^{y} + x\frac{d}{dx} (e^{y}) =y+ x\frac{ d }{dx }(y)\]now chain rule on exponential :
\[e^{y} + x \frac{d}{dy}(e^{y})\frac{dy}{dx}= y + x \frac{dy}{dx}\]derivative of exponential is itself :\[e^{y} + x e^{y}\frac{dy}{dx}= y + x \frac{dy}{dx}\]gather up like terms : \[\frac{dy}{dx}(x - x e^{y})= e^{y} - y\]nearly there : \[\frac{dy}{dx}= (e^{y} - y)/(x - x e^{y})\]whew ! I think that's right .... :-)

Answer 4

\[\Huge \checkmark \]

Answer 5

Thank you sir! :-)

Answer 6

Thank you so much :D