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Please explain geometrically.(https://www.desmos.com/calculator/ifwnwte5da) How to find tangent equation for the function $$y={\operatorname{mod}(x)}^{\frac12}$$ at the origin Differentiate once. We get slope undefined, i.e vertical line Double differentiate (as $dy/dx$ at that point is constant, its derivative is $0.$) we get $dy/dx=0.$ By solving We got the horizontal line too. Therefore we obtained both the tangents. But my professor says that that is not the correct process but equating sum of lowest degree terms to zero to get such tangent line is.

Also with the above method(mine) I obtained equations of tangents for the curve at the origin $$y^2=x^2(1+x+x^2)$$ I get $dy/dx=+1$ or $- 1$

By applying method given by my professor,i.e equating sum of low degree terms to zero $$y^2-x^2=0,$$ See the slopes,? here too $+1$ or $-1.$ (https://www.desmos.com/calculator/gx5zsqyctw) Please review my approach geometrically and explain its meaning of what I actually did.

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  • $\begingroup$ Assuming yo mean $\sqrt{|x|}$, Calculate left and right derivatives. RHD is $+\infty$ whereas LHD is $-\infty$ $\endgroup$
    – jonsno
    Commented Jan 5, 2018 at 14:48
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    $\begingroup$ Welcome to MSE! Try using mathjax to make your question more presentable. $\endgroup$ Commented Jan 5, 2018 at 14:49

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Unfortunately, what you did with the function $y = \sqrt{|x|}$ is invalid. If the first derivative doesn't exist (as you noted), you can't take the second derivative. In order to take a derivative of a function, a necessary condition is that the function be continuous; since the first derivative doesn't exist, it's not continuous, so you can't take a derivative of the first derivative, i.e., you can't take a second derivative.

Your method works with $y^2 = x^2(1+x+x^2)$ because the first derivatives here do exist (they are $\pm 1$ for the component curves, as you found), and because of one more factor in your favor. When you take two derivatives of each side, you get $2y\frac{d^2y}{dx^2} + 2(\frac{dy}{dx})^2 = 2 + 6x + 12x^2$; when you plug in $(x,y)=(0,0)$, you get an equation you can solve for $\frac{dy}{dx}$, but only because plugging in $y=0$ eliminates the unknown $\frac{d^2y}{dx^2}$ term. If you couldn't eliminate $\frac{d^2y}{dx^2}$, you'd merely have another unknown in the equation.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Commented Jan 9, 2018 at 16:35

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