Question

Find an equation of the tangent line to the bullet-nose curve y=|x|/sqrt(2−x^2) at the point (1,1) I think that square root is what is confusing me on this question

Answer 1

$\bf y=\cfrac{|x|}{\sqrt{2-x^2}}\qquad \boxed{|x|=\pm\sqrt{x^2}}\qquad y=\cfrac{\sqrt{x^2}}{\sqrt{2-x^2}}\\\\ -------------------------------\\\\ \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x\cdot \sqrt{2-x^2}~~-~~\sqrt{x^2}\cdot \frac{1}{2}(2-x^2)^{-\frac{1}{2}}\cdot -2x}{(\sqrt{2-x^2})^2}}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{x\sqrt{2-x^2}}{\sqrt{x^2}}~~+~~\frac{x\sqrt{x^2}}{\sqrt{2-x^2}}}{2-x^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{x(2-x^2)~~+~~x(x^2)}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}\\\\\\ \cfrac{dy}{dx}=\cfrac{\frac{2x\underline{-x^3+x^3}}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}\implies \cfrac{dy}{dx}=\cfrac{\frac{2x}{\sqrt{x^2}\sqrt{2-x^2}}}{2-x^2}$

$\bf \cfrac{dy}{dx}=\cfrac{2x}{\sqrt{x^2}\sqrt{2-x^2}(2-x^2)} \implies \boxed{\cfrac{dy}{dx}=\cfrac{2x}{|x|\sqrt{2-x^2}(2-x^2)}} \\\\\\ \left. \cfrac{dy}{dx} \right|_{1,1}\implies \cfrac{2(1)}{|1|\cdot \sqrt{2-1^2}(2-1^2)}\implies 2\\\\ -------------------------------\\\\ \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=2(x-1)\implies y-1=2x-2 \\\\\\ y=2x-1$