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Question Number 45776 by Sanjarbek last updated on 16/Oct/18

$$\mathit{y}=\mid \mathit{f}\left(\mathit{x}\right)\mid$$$${\mathit{y}}^{\prime }-?$$

Commented by maxmathsup by imad last updated on 16/Oct/18

$$forf\left(x\right)\ne 0wehave\mid f\left(x\right)\mid =\sqrt{f^2\left(x\right)}\Rightarrow \frac{d}{dx}(\mid f\left(x\right)\mid )=\frac{2f\left(x\right)f^{{}^{\prime }}\left(x\right)}{2\mid f\left(x\right)\mid }=\frac{f\left(x\right)f^{{}^{\prime }}\left(x\right)}{\mid f\left(x\right)\mid }$$$$=\xi \left(x\right).f^{{}^{\prime }}\left(x\right)with\xi \left(x\right)=1iff\left(x\right)>0and\xi \left(x\right)=-1iff\left(x\right)<0.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18

$$whenf\left(x\right)>0\mid f\left(x\right)\mid =+f\left(x\right)$$$$whenf\left(x\right)<0\mid f\left(x\right)\mid =-f\left(x\right)$$$$so\frac{dy}{dx}=f^{\prime }\left(x\right)f\left(x\right)>0$$$$\frac{dy}{dx}=-f^{\prime }x)f\left(x\right)<0$$$$so\frac{d\mid f\left(x\right)}{dx}doeznotexistplschecktheanswer$$

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