Question Number 121866 by danielasebhofoh last updated on 12/Nov/20
Answered by TANMAY PANACEA last updated on 12/Nov/20
![y=cos(x^x^x lnx) y=cosu→(dy/du)=−sinu=−sin(x^x^x lnx) u=x^x^x lnx u=pq (du/dx)=p(dq/dx)+q(dp/dx)=x^x^x ((1/x))+lnx×(dp/dx) red coloured to be found... p=x^x^x lnp=x^x lnx (1/p)×(dp/dx)=x^x ×(1/x)+lnx×(d/dx)(x^x ) (1/p)×(dp/dx)=x^(x−1) +lnx×x^x ×(1+lnx) (dp/dx)=x^x^x [x^(x−1) +lnx×x^x ×(1+lnx)] (du/dx)=[x^x^x ×(1/x)+lnx×x^x^x {x^(x−1) +lnx×x^x ×(1+lnx)}] so answer (dy/dx)=(dy/du)×(du/dx) =−sin(x^x^x lnx)×[x^x^x ×(1/x)+lnx×x^x^x {x^(x−1) +lnx×x^x (1+lnx)}] ★★ k=x^x →lnk=xlnx (1/k)×(dk/dx)=x×(1/x)+lnx=(dk/dx)=x^x (1+lnx)★★](https://www.tinkutara.com/question/Q121868.png)
$${y}={cos}\left({x}^{{x}^{{x}} } {lnx}\right) \\ $$$${y}={cosu}\rightarrow\frac{{dy}}{{du}}=−{sinu}=−{sin}\left({x}^{{x}^{{x}} } {lnx}\right) \\ $$$${u}={x}^{{x}^{{x}} } {lnx} \\ $$$${u}={pq} \\ $$$$\frac{{du}}{{dx}}={p}\frac{{dq}}{{dx}}+{q}\frac{{dp}}{{dx}}={x}^{{x}^{{x}} } \left(\frac{\mathrm{1}}{{x}}\right)+{lnx}×\frac{{dp}}{{dx}} \\ $$$$\boldsymbol{{red}}\:\boldsymbol{{coloured}}\:{to}\:{be}\:{found}… \\ $$$${p}={x}^{{x}^{{x}} } \\ $$$${lnp}={x}^{{x}} {lnx} \\ $$$$\frac{\mathrm{1}}{{p}}×\frac{{dp}}{{dx}}={x}^{{x}} ×\frac{\mathrm{1}}{{x}}+{lnx}×\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}} \right) \\ $$$$\frac{\mathrm{1}}{{p}}×\frac{{dp}}{{dx}}={x}^{{x}−\mathrm{1}} +{lnx}×{x}^{{x}} ×\left(\mathrm{1}+{lnx}\right) \\ $$$$\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \left[\boldsymbol{{x}}^{\boldsymbol{{x}}−\mathrm{1}} +\boldsymbol{{lnx}}×\boldsymbol{{x}}^{\boldsymbol{{x}}} ×\left(\mathrm{1}+\boldsymbol{{lnx}}\right)\right] \\ $$$$\frac{\boldsymbol{{du}}}{\boldsymbol{{dx}}}=\left[\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } ×\frac{\mathrm{1}}{\boldsymbol{{x}}}+\boldsymbol{{lnx}}×\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \left\{\boldsymbol{{x}}^{\boldsymbol{{x}}−\mathrm{1}} +\boldsymbol{{lnx}}×\boldsymbol{{x}}^{\boldsymbol{{x}}} ×\left(\mathrm{1}+\boldsymbol{{lnx}}\right)\right\}\right] \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\frac{\boldsymbol{{dy}}}{\boldsymbol{{du}}}×\frac{\boldsymbol{{du}}}{\boldsymbol{{dx}}} \\ $$$$=−\boldsymbol{{sin}}\left(\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \boldsymbol{{lnx}}\right)×\left[\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } ×\frac{\mathrm{1}}{\boldsymbol{{x}}}+\boldsymbol{{lnx}}×\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}} } \left\{\boldsymbol{{x}}^{\boldsymbol{{x}}−\mathrm{1}} +\boldsymbol{{lnx}}×\boldsymbol{{x}}^{\boldsymbol{{x}}} \left(\mathrm{1}+\boldsymbol{{lnx}}\right)\right\}\right] \\ $$$$\bigstar\bigstar\:{k}={x}^{{x}} \rightarrow{lnk}={xlnx}\: \\ $$$$\frac{\mathrm{1}}{{k}}×\frac{{dk}}{{dx}}={x}×\frac{\mathrm{1}}{{x}}+{lnx}=\frac{\boldsymbol{{dk}}}{\boldsymbol{{dx}}}=\boldsymbol{{x}}^{\boldsymbol{{x}}} \left(\mathrm{1}+\boldsymbol{{lnx}}\right)\bigstar\bigstar \\ $$$$ \\ $$
Commented by danielasebhofoh last updated on 12/Nov/20
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Answered by Dwaipayan Shikari last updated on 12/Nov/20
$$\frac{{d}}{{dx}}\left({cos}\left({x}^{{x}^{{x}} } {logx}\right)\right) \\ $$$$=−{sin}\left({x}^{{x}^{{x}} } {logx}\right)\left(\frac{\mathrm{1}}{{x}}.{x}^{{x}^{{x}} } +\frac{{d}\left({x}^{{x}^{{x}} } \right)}{{dx}}\right) \\ $$$$=−{x}^{{x}^{{x}} } {sin}\left({x}^{{x}^{{x}} } {logx}\right)\left(\frac{\mathrm{1}}{{x}}+{x}^{{x}} \left({log}^{\mathrm{2}} {x}+{logx}\right)+{x}^{{x}−\mathrm{1}} \right) \\ $$$${f}\left({x}\right)={x}^{{x}^{{x}} } \\ $$$${log}\left({f}\left({x}\right)\right)={x}^{{x}} {logx} \\ $$$$\frac{{f}'\left({x}\right)}{{f}\left({x}\right)}={x}^{{x}} \left({logx}+\mathrm{1}\right){logx}+{x}^{{x}−\mathrm{1}} \\ $$$${f}'\left({x}\right)={x}^{{x}^{{x}} } .{x}^{{x}} \left({logx}+\mathrm{1}\right){logx}+{x}^{{x}−\mathrm{1}} .{x}^{{x}^{{x}} } \\ $$