Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

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

Answered by Joel578 last updated on 20/Nov/19

∣x∣ = (√x^2 ) f(x) = (√((x^2 − 3)^2 )) f ′(x) = (1/(2(√((x^2 − 3)^2 )))) . 2(x^2 − 3) . 2x = ((2x(x^2 − 3))/(√((x^2 − 3)^2 ))) = ((2x(x^2 − 3))/(∣x^2 − 3∣))

$$\mid{x}\mid\:=\:\sqrt{{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)\:=\:\sqrt{\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${f}\:'\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)^{\mathrm{2}} }}\:.\:\mathrm{2}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)\:.\:\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)}{\sqrt{\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)}{\mid{x}^{\mathrm{2}} \:−\:\mathrm{3}\mid} \\ $$

Commented by MJS last updated on 20/Nov/19

(a/(∣a∣))=((∣a∣)/a)=sign a usually sign 0 :=1 ⇒ we can write f′(x)=2x sign (x^2 −3)

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

Commented by Joel578 last updated on 21/Nov/19

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

Answered by MJS last updated on 20/Nov/19

∣x^2 −3∣=x^2 −3∧x^2 ≥3 ∨ ∣x^2 −3∣=3−x^2 ∧x^2 <3 f(x)= { ((3−x^2 ∧−(√3)≤x≤(√3))),((x^2 −3∧(x<−(√3)∨x>(√3)))) :} f′(x)= { ((−2x∧−(√3)≤x≤(√3))),((2x∧(x<−(√3)∨x>(√3)))) :}

$$\mid{x}^{\mathrm{2}} −\mathrm{3}\mid={x}^{\mathrm{2}} −\mathrm{3}\wedge{x}^{\mathrm{2}} \geqslant\mathrm{3}\:\vee\:\mid{x}^{\mathrm{2}} −\mathrm{3}\mid=\mathrm{3}−{x}^{\mathrm{2}} \wedge{x}^{\mathrm{2}} <\mathrm{3} \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{3}−{x}^{\mathrm{2}} \wedge−\sqrt{\mathrm{3}}\leqslant{x}\leqslant\sqrt{\mathrm{3}}}\\{{x}^{\mathrm{2}} −\mathrm{3}\wedge\left({x}<−\sqrt{\mathrm{3}}\vee{x}>\sqrt{\mathrm{3}}\right)}\end{cases} \\ $$$${f}'\left({x}\right)=\begin{cases}{−\mathrm{2}{x}\wedge−\sqrt{\mathrm{3}}\leqslant{x}\leqslant\sqrt{\mathrm{3}}}\\{\mathrm{2}{x}\wedge\left({x}<−\sqrt{\mathrm{3}}\vee{x}>\sqrt{\mathrm{3}}\right)}\end{cases} \\ $$

Commented by aliesam last updated on 20/Nov/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

f(x) = ∣g(x)∣ f ′(x) = ((g(x))/(∣g(x)∣)) ∙ g^′ (x) let g(x) = x^2 −3 ⇒ g^′ (x) = 2x f^′ (x) = ((x^2 − 3)/(∣x^2 − 3∣)) ∙ 2x = ((2x^3 −6x)/(∣x^2 − 3∣))

$${f}\left({x}\right)\:=\:\mid{g}\left({x}\right)\mid \\ $$$${f}\:'\left({x}\right)\:=\:\frac{{g}\left({x}\right)}{\mid{g}\left({x}\right)\mid}\:\centerdot\:{g}\:^{'} \left({x}\right) \\ $$$${let}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:−\mathrm{3}\:\:\Rightarrow\:{g}\:^{'} \left({x}\right)\:=\:\mathrm{2}{x} \\ $$$${f}\:^{'} \left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:−\:\mathrm{3}}{\mid{x}^{\mathrm{2}} \:−\:\mathrm{3}\mid}\:\centerdot\:\mathrm{2}{x}\:=\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:−\mathrm{6}{x}}{\mid{x}^{\mathrm{2}} \:−\:\mathrm{3}\mid} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com