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Question Number 74235 by aliesam last updated on 20/Nov/19

Answered by Joel578 last updated on 20/Nov/19

$$\mid x\mid =\sqrt{x^2}$$$$f\left(x\right)=\sqrt{{(x^2-3)}^2}$$$$f{}^{\prime }\left(x\right)=\frac{1}{2\sqrt{{(x^2-3)}^2}}.2(x^2-3).2x$$$$=\frac{2x(x^2-3)}{\sqrt{{(x^2-3)}^2}}$$$$=\frac{2x(x^2-3)}{\mid x^2-3\mid }$$

Commented by MJS last updated on 20/Nov/19

$$\frac{a}{\mid a\mid }=\frac{\mid a\mid }{a}=\mathrm{sign}a$$$$\mathrm{usually}\mathrm{sign}0:=1$$$$\Rightarrow \mathrm{we}\mathrm{can}\mathrm{write}{f}^{\prime }\left(x\right)=2x\mathrm{sign}(x^2-3)$$

Commented by Joel578 last updated on 21/Nov/19

$$thankyouSir$$

Answered by MJS last updated on 20/Nov/19

$$\mid x^2-3\mid =x^2-3\wedge x^2⩾3\vee \mid x^2-3\mid =3-x^2\wedge x^2<3$$$$f\left(x\right)=\{\begin{array}{l}3-x^2\wedge -\sqrt{3}⩽x⩽\sqrt{3}\\ x^2-3\wedge (x<-\sqrt{3}\vee x>\sqrt{3})\end{array}$$$$f^{\prime }\left(x\right)=\{\begin{array}{l}-2x\wedge -\sqrt{3}⩽x⩽\sqrt{3}\\ 2x\wedge (x<-\sqrt{3}\vee x>\sqrt{3})\end{array}$$

Commented by aliesam last updated on 20/Nov/19

$$thankyousir$$

Answered by arkanmath7@gmail.com last updated on 21/Nov/19

$$f\left(x\right)=\mid g\left(x\right)\mid$$$$f{}^{\prime }\left(x\right)=\frac{g\left(x\right)}{\mid g\left(x\right)\mid }\cdot g{}^{{}^{\prime }}\left(x\right)$$$$letg\left(x\right)=x^2-3\Rightarrow g{}^{{}^{\prime }}\left(x\right)=2x$$$$f{}^{{}^{\prime }}\left(x\right)=\frac{x^2-3}{\mid x^2-3\mid }\cdot 2x=\frac{2x^3-6x}{\mid x^2-3\mid }$$

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