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Question Number 5587 by sanusihammed last updated on 21/May/16
![Please i need your help. if y = [tanx]^([tanx]^([tanx]) ) . find dy/dx at Π/4](Q5587.png)
$${Please}\:{i}\:{need}\:{your}\:{help}. \\ $$$$ \\ $$$${if}\:\:\:{y}\:\:=\:\:\left[{tanx}\right]^{\left[{tanx}\right]^{\left[{tanx}\right]} } \:\:\:.\:\:\:{find}\:\:{dy}/{dx}\:\:{at}\:\:\Pi/\mathrm{4} \\ $$
Commented by prakash jain last updated on 21/May/16
![[tan x]=greatest integer (floor function)? If not y=(tan x)^((tan x)^((tan x)) ) ln y=(tan x)^((tan x)) ln tan x ln ln y=tan xln (tan x)+ln (ln tan x) differentiating using chain rule (1/(ln y))∙(1/y)∙(dy/dx)=sec^2 xln (tan x)+tan x(1/(tan x))sec^2 x +(1/(ln (tan x)))∙(1/(tan x))∙sec^2 x (dy/dx)=yln y(sec^2 x)(1+ln tan x+(1/(tan x∙ln tan x))) (dy/dx)=(tan x)^((tan x)^((tan x)) ) ∙(tan x)^((tan x)) ln tan x(sec^2 x)(1+ln tan x+(1/(tan x∙ln tan x)))](Q5600.png)
$$\left[\mathrm{tan}\:{x}\right]={greatest}\:{integer}\:\left({floor}\:{function}\right)? \\ $$$$\mathrm{If}\:\mathrm{not}\: \\ $$$${y}=\left(\mathrm{tan}\:{x}\right)^{\left(\mathrm{tan}\:{x}\right)^{\left(\mathrm{tan}\:{x}\right)} } \\ $$$$\mathrm{ln}\:{y}=\left(\mathrm{tan}\:{x}\right)^{\left(\mathrm{tan}\:{x}\right)} \mathrm{ln}\:\mathrm{tan}\:{x} \\ $$$$\mathrm{ln}\:\mathrm{ln}\:{y}=\mathrm{tan}\:{x}\mathrm{ln}\:\left(\mathrm{tan}\:{x}\right)+\mathrm{ln}\:\left(\mathrm{ln}\:\mathrm{tan}\:{x}\right) \\ $$$${differentiating}\:{using}\:{chain}\:{rule} \\ $$$$\frac{\mathrm{1}}{\mathrm{ln}\:{y}}\centerdot\frac{\mathrm{1}}{{y}}\centerdot\frac{\mathrm{d}{y}}{\mathrm{d}{x}}=\mathrm{sec}^{\mathrm{2}} {x}\mathrm{ln}\:\left(\mathrm{tan}\:{x}\right)+\mathrm{tan}\:{x}\frac{\mathrm{1}}{\mathrm{tan}\:{x}}\mathrm{sec}^{\mathrm{2}} {x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{ln}\:\left(\mathrm{tan}\:{x}\right)}\centerdot\frac{\mathrm{1}}{\mathrm{tan}\:{x}}\centerdot\mathrm{sec}^{\mathrm{2}} {x} \\ $$$$\frac{{dy}}{{dx}}={y}\mathrm{ln}\:{y}\left(\mathrm{sec}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+\mathrm{ln}\:\mathrm{tan}\:{x}+\frac{\mathrm{1}}{\mathrm{tan}\:{x}\centerdot\mathrm{ln}\:\mathrm{tan}\:{x}}\right) \\ $$$$\frac{{dy}}{{dx}}=\left(\mathrm{tan}\:{x}\right)^{\left(\mathrm{tan}\:{x}\right)^{\left(\mathrm{tan}\:{x}\right)} } \centerdot\left(\mathrm{tan}\:{x}\right)^{\left(\mathrm{tan}\:{x}\right)} \mathrm{ln}\:\mathrm{tan}\:{x}\left(\mathrm{sec}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+\mathrm{ln}\:\mathrm{tan}\:{x}+\frac{\mathrm{1}}{\mathrm{tan}\:{x}\centerdot\mathrm{ln}\:\mathrm{tan}\:{x}}\right) \\ $$