Compute [tex]\frac{\mathrm dy}{\mathrm dx}[/tex] using implicit differentiation:
[tex]x^2-(xy)^{1/2}=2y^2+3x[/tex]
[tex]\dfrac{\mathrm d(x^2)}{\mathrm dx}-\dfrac{\mathrm d(xy)^{1/2}}{\mathrm dx}=\dfrac{\mathrm d(2y^2)}{\mathrm dx}+\dfrac{\mathrm d(3x)}{\mathrm dx}[/tex]
[tex]2x-\dfrac1{2(xy)^{1/2}}\left(\dfrac{\mathrm dx}{\mathrm dx}y+x\dfrac{\mathrm dy}{\mathrm dx}\right)=4y\dfrac{\mathrm dy}{\mathrm dx}+3[/tex]
[tex]2x-\dfrac1{2(xy)^{1/2}}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)=4y\dfrac{\mathrm dy}{\mathrm dx}+3[/tex]
[tex]2x-\dfrac12\left(\dfrac yx\right)^{1/2}-\dfrac12\left(\dfrac xy\right)^{1/2}\dfrac{\mathrm dy}{\mathrm dx}=4y\dfrac{\mathrm dy}{\mathrm dx}+3[/tex]
[tex]2x-3-\dfrac12\left(\dfrac yx\right)^{1/2}=\left(4y+\dfrac12\left(\dfrac xy\right)^{1/2}\right)\right)\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]4x-6-\left(\dfrac yx\right)^{1/2}=\left(8y+\left(\dfrac xy\right)^{1/2}\right)\right)\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4x-6-\left(\frac yx\right)^{1/2}}{8y+\left(\frac xy\right)^{1/2}}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4x(xy)^{1/2}-6(xy)^{1/2}-y}{8y(xy)^{1/2}+x}[/tex]
Plug in x = 4 and y = 1 to find that the tangent line's slope is 19/20.
