Suppose $xe^y-5y=2x+14-2e^{-2}$ and $y(-2)=-2$. How to find $y'(-2)$?

Question

Suppose $xe^y-5y=2x+14-2e^{-2}$ and $y(-2)=-2$. Find $y'(-2)$.

Just learning this and confused on how to solve, can someone show me the solution step by step?

Answer 1

HINT.-You have $$F(x,y)=xe^y-5y-2x+(2e^{-2}-14)=0$$ Then $$y'(x)=\frac{2-e^y}{xe^y-5}$$ $$y(-2)=-2\Rightarrow y'(-2)=\frac{e^{-2}-2}{2e^{-2}+5}$$You can verify that it is truly the asked derivative plotting the curve and the tangent at the point $(-2,-2)$ using this derivative as pente.

Answer 2

$$xe^y-5y=2x+14-2e^{-2}$$
Put $y = \ln t,$ $$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}

\frac{dy}{dt}=\frac1t$$

So substituting into 1

$xt-5\ln t=2x+14-2e^{-2}$

Differentiating

$t+xdt/dx-5tdt/dx=2

dt/dx=(2-t)/(x-5)

dt/dx=(2-e^y)/(x-5) dt/dx=(2-e^-2)/(-7)

dy/dt=1/t=1/e^y=e^2

dy/dx=e^2(2-e^-2)/-7

=-2(e^2-1)/7$ Merci