Question Number 63888 by Mikael last updated on 10/Jul/19
![y = log_2 [log_3 (log_5 x)] y = ?](Q63888.png)
$${y}\:=\:{log}_{\mathrm{2}} \left[{log}_{\mathrm{3}} \left({log}_{\mathrm{5}} {x}\right)\right] \\ $$$${y} \:=\:? \\ $$
Answered by Hope last updated on 10/Jul/19
![y=log_2 [log_3 (u)] →u=log_5 x x=5^u →(dx/du)=5^u ln5=xln5 y=log_2 [v]→v=log_3 u 3^v =u (du/dv)=3^v ln3=uln3=(log_5 x)ln3 y=log_2 v v=2^y (dv/dy)=2^y ln2=vln2=2^y ln2 now (dv/dy)×(du/dv)×(dx/du)=(2^y ln2)×(log_5 x ln3)×(xln5) (dx/dy)=(ln2×ln3×ln5)(xlog_5 x)×(2^y ) (dx/dy)=(ln2×ln3×ln5)×(xln_5 x)×2^(log_2 (log_3 (log_5 x))) (dy/dx)=(1/((ln2×ln3×ln5)))×(1/(xln_5 x))×(1/2^(log_2 (log_3 (log_5 x))) ) pls check](Q63890.png)
$${y}={log}_{\mathrm{2}} \left[{log}_{\mathrm{3}} \left({u}\right)\right]\:\rightarrow{u}={log}_{\mathrm{5}} {x} \\ $$$${x}=\mathrm{5}^{{u}} \:\:\rightarrow\frac{{dx}}{{du}}=\mathrm{5}^{{u}} {ln}\mathrm{5}={xln}\mathrm{5} \\ $$$${y}={log}_{\mathrm{2}} \left[{v}\right]\rightarrow{v}={log}_{\mathrm{3}} {u} \\ $$$$\mathrm{3}^{{v}} ={u} \\ $$$$\frac{{du}}{{dv}}=\mathrm{3}^{{v}} {ln}\mathrm{3}={uln}\mathrm{3}=\left({log}_{\mathrm{5}} {x}\right){ln}\mathrm{3} \\ $$$${y}={log}_{\mathrm{2}} {v} \\ $$$${v}=\mathrm{2}^{{y}} \\ $$$$\frac{{dv}}{{dy}}=\mathrm{2}^{{y}} {ln}\mathrm{2}={vln}\mathrm{2}=\mathrm{2}^{{y}} {ln}\mathrm{2} \\ $$$${now}\:\frac{{dv}}{{dy}}×\frac{{du}}{{dv}}×\frac{{dx}}{{du}}=\left(\mathrm{2}^{{y}} {ln}\mathrm{2}\right)×\left({log}_{\mathrm{5}} {x}\:{ln}\mathrm{3}\right)×\left({xln}\mathrm{5}\right) \\ $$$$\frac{{dx}}{{dy}}=\left({ln}\mathrm{2}×{ln}\mathrm{3}×{ln}\mathrm{5}\right)\left({xlog}_{\mathrm{5}} {x}\right)×\left(\mathrm{2}^{{y}} \right) \\ $$$$\frac{{dx}}{{dy}}=\left({ln}\mathrm{2}×{ln}\mathrm{3}×{ln}\mathrm{5}\right)×\left({xln}_{\mathrm{5}} {x}\right)×\mathrm{2}^{{log}_{\mathrm{2}} \left({log}_{\mathrm{3}} \left({log}_{\mathrm{5}} {x}\right)\right)} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\left({ln}\mathrm{2}×{ln}\mathrm{3}×{ln}\mathrm{5}\right)}×\frac{\mathrm{1}}{{xln}_{\mathrm{5}} {x}}×\frac{\mathrm{1}}{\mathrm{2}^{{log}_{\mathrm{2}} \left({log}_{\mathrm{3}} \left({log}_{\mathrm{5}} {x}\right)\right)} } \\ $$$${pls}\:{check} \\ $$
Commented by Mikael last updated on 10/Jul/19
$${thank}\:{you}\:{Sir} \\ $$
Answered by mr W last updated on 10/Jul/19
![y′=(1/(ln 2×log_3 (log_5 x)))×(1/(ln 3×log_5 x))×(1/(ln 5×x)) y′=(1/(ln 2×ln 3×ln 5×[log_3 (log_5 x)][log_5 x]x))](Q63896.png)
$${y}'=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}×\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{5}} \:{x}\right)}×\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{3}×\mathrm{log}_{\mathrm{5}} \:{x}}×\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{5}×{x}} \\ $$$${y}'=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}×\mathrm{ln}\:\mathrm{3}×\mathrm{ln}\:\mathrm{5}×\left[\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{5}} \:{x}\right)\right]\left[\mathrm{log}_{\mathrm{5}} \:{x}\right]{x}} \\ $$
Commented by Mikael last updated on 10/Jul/19
$${Thank}\:{you}\:{Sir}. \\ $$