What is the slope of the tangent line of (xy^2-3xy)/(sqrt(yx)-xy) =C , where C is an arbitrary constant, at (2,5)?

1 Answer

-0.653281301

Explanation:

Given implicit function \frac{xy^2-3xy}{\sqrt{yx}-xy}=C

Differentiating the above function wrt x on both the sides we get

\frac{(\sqrt{yx}-xy)\frac{d}{dx}(xy^2-3xy)-(xy^2-3xy)\frac{d}{dx}(\sqrt{yx}-xy)}{(\sqrt{yx}-xy)^2}=0

(\sqrt{yx}-xy)(2xyy'+y^2-3xy'-3y)-(xy^2-3xy)(1/2\sqrt{x/y}y'+1/2\sqrt{y/x})=0\quad (\forall \ xy\ne0,1)

(2(\sqrt{xy}-xy)(2xy-3x)-\sqrt{x/y}(x^2y-3xy))y'=\sqrt{y/x}(x^2y-3xy)-2(\sqrt{xy}-xy)(y^2-3y)

y'=\frac{\sqrt{y/x}(x^2y-3xy)-2(\sqrt{xy}-xy)(y^2-3y)}{2(\sqrt{xy}-xy)(2xy-3x)-\sqrt{x/y}(x^2y-3xy)}

Now, setting x=2, y=5 in above equation, we get the slope y' of the tangent line

y'=\frac{\sqrt{5/2}(2^2\cdot5-3(2)(5))-2(\sqrt{2\cdot 5}-2\cdot5)(5^2-3\cdot 5)}{2(\sqrt{2\cdot 5}-2\cdot 5)(2\cdot 2\cdot 5-3\cdot 2)-\sqrt{2/5}(2^2\cdot5-3\cdot 2\cdot 5)}

y'=120.9430585/-185.1316702

y'=-0.653281301