[tex]xe^y+4\ln y=x^2[/tex]
Differentiate both sides with respect to x, assuming y = y(x).
[tex]\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}[/tex]
[tex]\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x[/tex]
[tex]\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x[/tex]
[tex]e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x[/tex]
Solve for dy/dx :
[tex]e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x[/tex]
[tex]\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}[/tex]
If y ≠ 0, we can write
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}[/tex]
At the point (1, 1), the derivative is
[tex]\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}[/tex]
