Answer to Question #302720 in Calculus for freebandz

Question

1) Find the tangent line to the graph given by x^2(x^2+y^2)=y^2 at the points(√2/2, √2/2)

Answer 1

Differentiate both sides with respect to $x$

$$(x^2(x^2+y^2))'=(y^2)'$$

Use the Chain Rule

$$4x^3+2xy^2+2x^2yy'=2yy'$$

Solve for $y'$

$$y'=\dfrac{4x^3+2xy^2}{2y(1-x^2)}$$

Point $(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})$

$$y'(\dfrac{\sqrt{2}}{2})=\dfrac{4(\dfrac{\sqrt{2}}{2})^3+2(\dfrac{\sqrt{2}}{2})(\dfrac{\sqrt{2}}{2})^2}{2(\dfrac{\sqrt{2}}{2})(1-(\dfrac{\sqrt{2}}{2})^2)}=2$$

The tangent line to the graph 

$$y-\dfrac{\sqrt{2}}{2}=2(x-\dfrac{\sqrt{2}}{2})$$

$$y=2x-\dfrac{\sqrt{2}}{2}$$