Question-105388

Question Number 105388 by mohammad17 last updated on 28/Jul/20

Answered by john santu last updated on 28/Jul/20

let u = log _2 (x^3 −1) (du/dx) = ((3x^2 )/((x^3 −1)ln 2)) (1)y = u^4 ⇒(dy/du) = 4u^3 (dy/dx) = (dy/du)×(du/dx) = 4u^3 ×((3x^2 )/((x^3 −1)ln 2)) (dy/dx) = ((12x^2 (log _2 (x^3 −1))^3 )/((x^3 −1)ln 2))

$${let}\:{u}\:=\:\mathrm{log}\:_{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\frac{{du}}{{dx}}\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} \:}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\mathrm{1}\right){y}\:=\:{u}^{\mathrm{4}} \:\Rightarrow\frac{{dy}}{{du}}\:=\:\mathrm{4}{u}^{\mathrm{3}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}×\frac{{du}}{{dx}}\:=\:\mathrm{4}{u}^{\mathrm{3}} ×\frac{\mathrm{3}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{3}} −\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{12}{x}^{\mathrm{2}} \left(\mathrm{log}\:_{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{3}} }{\left({x}^{\mathrm{3}} −\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Jul/20

y=(log_2 (x^3 −1))^4 y=(((log(x^3 −1))^4 )/(log2)) (dy/dx)=(4/(log2))(log(x^3 −1))^3 ((3x^2 )/(x^3 −1))=((12)/(log2))(log(x^3 −1))(1/(x^3 −1))

$${y}=\left({log}_{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{4}} \\ $$$${y}=\frac{\left({log}\left({x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{4}} }{{log}\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}}{{log}\mathrm{2}}\left({log}\left({x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{3}} \frac{\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} −\mathrm{1}}=\frac{\mathrm{12}}{{log}\mathrm{2}}\left({log}\left({x}^{\mathrm{3}} −\mathrm{1}\right)\right)\frac{\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$

Answered by mathmax by abdo last updated on 28/Jul/20

y(x) =(log_2 (x^3 −1))^4 ⇒y(x) ={((ln(x^3 −1))/(ln2))}^4 =(1/((ln2)^4 ))(ln(x^3 −1))^4 ⇒y^′ (x) =(1/((ln2)^4 ))×4×((3x^2 )/(x^3 −1))(ln(x^3 −1))^3 =((12x^2 )/((ln2)^4 (x^3 −1))) (ln(x^3 −1))^3

$$\mathrm{y}\left(\mathrm{x}\right)\:=\left(\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{4}} \:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)\:=\left\{\frac{\mathrm{ln}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)}{\mathrm{ln2}}\right\}^{\mathrm{4}} \:=\frac{\mathrm{1}}{\left(\mathrm{ln2}\right)^{\mathrm{4}} }\left(\mathrm{ln}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{ln2}\right)^{\mathrm{4}} }×\mathrm{4}×\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}\left(\mathrm{ln}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{3}} \\ $$$$=\frac{\mathrm{12x}^{\mathrm{2}} }{\left(\mathrm{ln2}\right)^{\mathrm{4}} \left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)}\:\left(\mathrm{ln}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{1}\right)\right)^{\mathrm{3}} \\ $$

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