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Question Number 134127 by bobhans last updated on 28/Feb/21

$$\mathrm{Given}\mid \mathrm{f}\left(\mathrm{x}\right)\mid ={\mathrm{x}}^2\mathrm{for}-1<\mathrm{x}<1$$ $$\mathrm{find}\mathrm{f}{}^{\prime }\left(0\right).$$

Commented byEDWIN88 last updated on 28/Feb/21

$$\mathrm{does}\mathrm{not}\mathrm{exist}$$

Commented byEDWIN88 last updated on 28/Feb/21

Commented byMJS_new last updated on 28/Feb/21

$$\mathrm{proper}\mathrm{maths}:$$ $${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{try}\mathrm{with}$$ $$f\left(x\right)=2x\mathrm{\forall }x\in \mathbb{R}$$ $$\mathrm{this}\mathrm{simply}\mathrm{means},{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{assigning}x\to 2x$$ $$\mathrm{or}y=2x$$ $$\mid f\left(x\right)\mid =2x\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{function}\mathrm{anymore}$$ $$\mid y\mid =2x\iff y=\pm 2x\wedge x⩾0$$ $$\mathrm{we}\mathrm{can}\mathrm{split}\mathrm{the}\mathrm{branches}:$$ $$y_1=-2x\wedge x⩾0$$ $$y_2=2x\wedge x⩾0$$ $$\frac{{dy}_1}{dx}\mathrm{and}\frac{{dy}_2}{dx}\mathrm{both}\mathrm{exist}\mathrm{but}\frac{dy}{dx}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$ $$$$ $$\mathrm{now}$$ $$\mid f\left(x\right)\mid =x^2\mathrm{\forall }x\in \mathbb{R}$$ $$y=\pm x^2\wedge x^2⩾0\mathrm{but}{x}^2⩾0\mathrm{\forall }x\in \mathbb{R}$$ $$\mathrm{again}\mathrm{we}\mathrm{have}2\mathrm{branches}:$$ $$y_1=-x^2\mathrm{\forall }x\in \mathbb{R}$$ $$y_2=x^2\mathrm{\forall }x\in \mathbb{R}$$ $$\mathrm{again}\frac{{dy}_1}{dx}\mathrm{and}\frac{{dy}_2}{dx}\mathrm{exist}\mathrm{but}\frac{dy}{dx}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$ $$\mathrm{anyway}\mathrm{for}x=0\mathrm{we}\mathrm{have}\frac{{dy}_1}{dx}=\frac{{dy}_2}{dx}=0$$ $$\mathrm{but}\mathrm{this}\mathrm{still}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{mean}\mathrm{that}\frac{dy}{dx}=0\mathrm{for}x=0$$

Answered by EDWIN88 last updated on 28/Feb/21

$$\mid \mathrm{y}\mid ={\mathrm{x}}^2\Rightarrow \sqrt{{\mathrm{y}}^2}={\mathrm{x}}^2;{\mathrm{y}}^2={\mathrm{x}}^4$$ $$2\mathrm{y}{\mathrm{y}}^{\prime }={4\mathrm{x}}^3,{\mathrm{y}}^{\prime }=\frac{{2\mathrm{x}}^3}{\sqrt{\mathrm{y}}}=\frac{0}{0}\left(\mathrm{does}\mathrm{not}\mathrm{exist}\right)$$

Commented bymr W last updated on 28/Feb/21

$${\mathrm{y}}^{\prime }=\frac{{2\mathrm{x}}^3}{\sqrt{\mathrm{y}}}=\frac{2x^3}{x^2}=2x$$ $$f^{\prime }\left(0\right)=0$$

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