# Find the slope of the tangent line to the curve.

So I am trying to find the slope of the tangent line to the curve

$$\sqrt{4x+2y} + \sqrt{xy} = \sqrt{38} + \sqrt{24}.$$

at the point $(8,3)$.

I ended up implicitly differentiating and getting $\mathrm dy/\mathrm dx$ on one side and then plugging $(8,3)$ in place of $x$ and $y$ respectively.

I ended up getting a slope of $.644341944$; however, that is incorrect. I am unsure what I did wrong now and how to properly do then problem now then :/

Thanks for the help.

• Please show the work you have done so we can help spot where you may have made a mistake – Mufasa Jun 1 '14 at 19:39

$$\sqrt{4x+2y}+\sqrt{xy}=\sqrt{38}+\sqrt{24}\implies$$
$$\frac2{\sqrt{4x+2y}}dx+\frac1{\sqrt{4x+2y}}dy+\frac y{2\sqrt{xy}}dx+\frac x{2\sqrt{xy}}dy=0\implies$$
$$\left(\frac1{\sqrt{4x+2y}}+\frac12\sqrt\frac xy\right)dy=-\left(\frac2{\sqrt{4x+2y}}+\frac12\sqrt\frac yx\right)\implies$$
$$\frac{dy}{dx}=-\frac{\left(\frac2{\sqrt{4x+2y}}+\frac12\sqrt\frac yx\right)}{\left(\frac1{\sqrt{4x+2y}}+\frac12\sqrt\frac xy\right)}\implies$$
$$\frac{dy}{dx}(8,3)=-\frac{\frac2{\sqrt{38}}+\frac14\sqrt\frac32}{\frac1{\sqrt{38}}+\sqrt\frac23}$$