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=\cfracx{√(2-x^2)}\qquad \boxed=\pm√(x^2)\qquad y=\cfrac{√(x^2)}{√(2-x^2)}\\\\ -------------------------------\\\\ \cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{(1)/(2)(x^2)^{-(1)/(2)}\cdot 2x\cdot √(2-x^2)~~-~~√(x^2)\cdot (1)/(2)(2-x^2)^{-(1)/(2)}\cdot -2x}{(√(2-x^2))^2}}`


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


`\bf \cfrac{dy}{dx}=\cfrac{2x}{√(x^2)√(2-x^2)(2-x^2)} \implies \boxed{\cfrac{dy}{dx}=\cfrac{2x}} \\\\\\ \left. \cfrac{dy}{dx} \right|_(1,1)\implies \cfrac{2(1)}\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`