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Question Number 95339 by mathocean1 last updated on 24/May/20

f(x)=∣2x+3∣ f ′(x)=...?

$$\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{2x}+\mathrm{3}\mid \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=...? \\ $$

Answered by john santu last updated on 24/May/20

f(x) = (√((2x+3)^2 )) f ′(x) = ((2(2x+3)^1 .2)/(2(√((2x+3)^2 )))) f ′(x) = ((4x+6)/(∣2x+3∣))

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\left(\mathrm{2x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{2}\left(\mathrm{2x}+\mathrm{3}\right)^{\mathrm{1}} .\mathrm{2}}{\mathrm{2}\sqrt{\left(\mathrm{2x}+\mathrm{3}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{4x}+\mathrm{6}}{\mid\mathrm{2x}+\mathrm{3}\mid} \\ $$

Answered by prakash jain last updated on 24/May/20

x=(y/2)−(3/2)⇒y=2(x+(3/2)) ∣2x−3∣=∣y∣ ((df(x))/dx)=(d/dy)∣y∣×(dy/dx)=(y/(∣y∣))×2 =((2(x+(3/2)))/(∣2(x+(3/2))∣))×2=((4x+6)/(∣2x+3∣))

$${x}=\frac{{y}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{y}=\mathrm{2}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\mid\mathrm{2}{x}−\mathrm{3}\mid=\mid{y}\mid \\ $$$$\frac{{df}\left({x}\right)}{{dx}}=\frac{{d}}{{dy}}\mid{y}\mid×\frac{{dy}}{{dx}}=\frac{{y}}{\mid{y}\mid}×\mathrm{2} \\ $$$$=\frac{\mathrm{2}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mid\mathrm{2}\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)\mid}×\mathrm{2}=\frac{\mathrm{4}{x}+\mathrm{6}}{\mid\mathrm{2}{x}+\mathrm{3}\mid} \\ $$

Commented by john santu last updated on 24/May/20

yes. your are right

$$\mathrm{yes}.\:\mathrm{your}\:\mathrm{are}\:\mathrm{right} \\ $$

Answered by  M±th+et+s last updated on 24/May/20

by first principle lim_(h→0) ((∣2(x+h)+3∣−∣2x+3∣)/( h)).((∣2(x+h)+3∣+∣2x+3∣)/(∣2(x+h)+3∣+∣2x+3∣)) lim_(h→0) ((∣(2x+3)+2h∣^2 −∣2x+3∣^2 )/(h(∣2(x+h)+3∣+∣2x+3∣))) lim_(h→0) (((2x+3)^2 +4(2x+3)h+4h^2 −(2x+3)^2 )/(h(∣(2x+3)+2h∣+∣2x+3∣))) lim_(h→0) ((4h(2x+3+h))/(h(∣(2x+3)+2h∣+∣2x+3∣))) lim_(x→0) ((4(2x+3+h))/(∣2x+3+h∣+∣2x+3∣))=((4x+6)/(∣2x+3∣))

$${by}\:{first}\:{principle} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{\mid\mathrm{2}\left({x}+{h}\right)+\mathrm{3}\mid−\mid\mathrm{2}{x}+\mathrm{3}\mid}{\:{h}}.\frac{\mid\mathrm{2}\left({x}+{h}\right)+\mathrm{3}\mid+\mid\mathrm{2}{x}+\mathrm{3}\mid}{\mid\mathrm{2}\left({x}+{h}\right)+\mathrm{3}\mid+\mid\mathrm{2}{x}+\mathrm{3}\mid} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{\mid\left(\mathrm{2}{x}+\mathrm{3}\right)+\mathrm{2}{h}\mid^{\mathrm{2}} −\mid\mathrm{2}{x}+\mathrm{3}\mid^{\mathrm{2}} }{{h}\left(\mid\mathrm{2}\left({x}+{h}\right)+\mathrm{3}\mid+\mid\mathrm{2}{x}+\mathrm{3}\mid\right)} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{2}{x}+\mathrm{3}\right){h}+\mathrm{4}{h}^{\mathrm{2}} −\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} }{{h}\left(\mid\left(\mathrm{2}{x}+\mathrm{3}\right)+\mathrm{2}{h}\mid+\mid\mathrm{2}{x}+\mathrm{3}\mid\right)} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{4}{h}\left(\mathrm{2}{x}+\mathrm{3}+{h}\right)}{{h}\left(\mid\left(\mathrm{2}{x}+\mathrm{3}\right)+\mathrm{2}{h}\mid+\mid\mathrm{2}{x}+\mathrm{3}\mid\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{4}\left(\mathrm{2}{x}+\mathrm{3}+{h}\right)}{\mid\mathrm{2}{x}+\mathrm{3}+{h}\mid+\mid\mathrm{2}{x}+\mathrm{3}\mid}=\frac{\mathrm{4}{x}+\mathrm{6}}{\mid\mathrm{2}{x}+\mathrm{3}\mid} \\ $$$$ \\ $$

Commented by john santu last updated on 24/May/20

∣2(x+h)+3∣ = ∣(2x+3)+2h∣

$$\mid\mathrm{2}\left(\mathrm{x}+\mathrm{h}\right)+\mathrm{3}\mid\:=\:\mid\left(\mathrm{2x}+\mathrm{3}\right)+\mathrm{2h}\mid\: \\ $$

Commented by  M±th+et+s last updated on 24/May/20

yes i didn′t focus.thank you sir

$${yes}\:{i}\:{didn}'{t}\:{focus}.{thank}\:{you}\:{sir}\: \\ $$

Answered by mathmax by abdo last updated on 24/May/20

f(x)= { ((2x+3 if x>−(3/2) and f(−(3/2))=0 ⇒)),((−2x−3 if x<−(3/2) )) :} f^′ (x) = { ((2 if x>−(3/2))),((−2 if x<−(3/2))) :} rest studying derivability at x_0 =−(3/2) ....

$$\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{2x}+\mathrm{3}\:\mathrm{if}\:\mathrm{x}>−\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{and}\:\mathrm{f}\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0}\:\Rightarrow}\\{−\mathrm{2x}−\mathrm{3}\:\:\mathrm{if}\:\:\mathrm{x}<−\frac{\mathrm{3}}{\mathrm{2}}\:}\end{cases} \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)\:=\begin{cases}{\mathrm{2}\:\mathrm{if}\:\mathrm{x}>−\frac{\mathrm{3}}{\mathrm{2}}}\\{−\mathrm{2}\:\mathrm{if}\:\mathrm{x}<−\frac{\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{rest}\:\mathrm{studying}\:\mathrm{derivability}\:\mathrm{at}\:\mathrm{x}_{\mathrm{0}} =−\frac{\mathrm{3}}{\mathrm{2}}\:.... \\ $$

Answered by 1549442205 last updated on 25/May/20

we have f^2 (x)=4x^2 +12x+9⇒ 2f(x).f^′ (x)=8x+12⇒f^′ (x)=((8x+12)/(2f(x)))=((4x+6)/(∣2x+3∣))

$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{4x}^{\mathrm{2}} +\mathrm{12x}+\mathrm{9}\Rightarrow \\ $$$$\mathrm{2f}\left(\mathrm{x}\right).\mathrm{f}\:^{'} \left(\mathrm{x}\right)=\mathrm{8x}+\mathrm{12}\Rightarrow\mathrm{f}\:^{'} \left(\mathrm{x}\right)=\frac{\mathrm{8x}+\mathrm{12}}{\mathrm{2f}\left(\mathrm{x}\right)}=\frac{\mathrm{4x}+\mathrm{6}}{\mid\mathrm{2x}+\mathrm{3}\mid}\: \\ $$

Answered by mathmax by abdo last updated on 10/Oct/21

f(x)=(√((2x+3)^2 )) ⇒f^′ (x)=((2.2(2x+3))/(2(√((2x+3)^2 ))))=((4x+6)/(∣2x+3∣))

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{2x}+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{2}.\mathrm{2}\left(\mathrm{2x}+\mathrm{3}\right)}{\mathrm{2}\sqrt{\left(\mathrm{2x}+\mathrm{3}\right)^{\mathrm{2}} }}=\frac{\mathrm{4x}+\mathrm{6}}{\mid\mathrm{2x}+\mathrm{3}\mid} \\ $$

Commented by prakash jain last updated on 10/Oct/21

how are you, sir

$$\mathrm{how}\:\mathrm{are}\:\mathrm{you},\:\mathrm{sir} \\ $$

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