# y= log (log 8 (log 7 (log 6 (log 5 (x))))) find the domain of dy/dx

y= log (log 8 (log 7 (log 6 (log 5 (x))))) find the domain of dy/dx

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𝑦 x log9(log8(log7(log6(log5 𝑥)))) y is xxx xxxxxxxxx xxx xx have to xxxx xxx xxxxxxxxxxx 𝑑𝑦

𝑑𝑥

𝑑𝑦

𝑑𝑥 = 𝑦′ = 1log8(log7(log6(log5 𝑥xxx ∙ xx x xxx [log8(log7(log6(log5 𝑥)))]′ = xxxxxxxxxxxxxxxxxxxx 𝑥xxx ∙ ln x xxx xxxxxxxxxxxxxxx 𝑥)) xxx ln 8 xxxxxxxxxxxxxxx 𝑥xxx′ x 1log8(log7(log6(log5 𝑥))) xxx xx x xxx xxxxxxxxxxxxxxx 𝑥)) xxx xx 8 xxx xxxxxxxxxx 𝑥) xxx xx x xxxxxxxxxx 𝑥)]′ x xxxxxxxxxxxxxxxxxxxx 𝑥xxx ∙ xx 9 ∙ xxxxxxxxxxxxxxx 𝑥)) xxx ln x xxx xxxxxxxxxx 𝑥x xxx ln 7

xxxxxxxxx 𝑥 ∙ ln 6 xxxxx 𝑥x′ = xxxxxxxxxxxxxxxxxxxx 𝑥xxx xxx xx 9 ∙ 1log7(log6(log5 𝑥)) ∙ xx x ∙ 1log6(log5 𝑥x ∙ xx 7 ∙ xxxxx 𝑥 xxx ln 6

xxxxxx

𝑥 ∙ xx 5 xxx in xxxxx to find xxx domain of this xxxxxxxxxx xx xxxx multiple conditions. xxxxxxx all xxxxxxxxx xx xxxxxxxxxx must be xxxxxxx than xxxxx

xxx xxxxxxxxxxxxxx 𝑥)) > 0 xxxxxxxxxxxxxxx 𝑥xx > xx xxxxxxxxx 𝑥) > 1 xxxxxxxxxx 𝑥x > xx log5 𝑥 > 6 5log5 𝑥 > 56 𝑥 > xx

xxx log6(log5 𝑥) > 0 6log6(log5 𝑥x > 60 log5 𝑥 > x 5log5 𝑥 > 51 𝑥

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Y is xxx xxxxxxxxx Now we xxxx to xxxx its xxxxxxxxxxx

Now in order xx find xxx domain xx this xxxxxxxxxx xx xxxx multiple conditions. Firslty all xxxxxxxxx xx xxxxxxxxxx xxxx be greater xxxx xxxxx

xxxxxxxxx xxx xxxxxxxxxxxx xxxx xx xxxxxxxxx xxxx xxxxx

xxxxxxxxx , x , x x are surely real xxx xxxxxxxxx xxxx zero.

The final domain xx xxx xxxxxxxxx is the xxxxxxxxx xx xxx conditions. The xxxxxx xxxxxxxx for the xxxxx x conditions is:

Because xx xxxxxxx than all other numbers xx xxxxxxxxxxx xx xxxxx words xx xxx x xxxx xx greater than xx xx xx than it xxxx be greater than .

xx must xxxxxxx xxxx xxx domain xxx values of x xxxx conditions xxxxx xxxxx xxxxxxx 0, 1, xx , are not in xxx xxxxxxxxx xxxxx xx xx xxxxx in xxxxxxxxx them. xxx only number we must exclude xx :

This xx the xxxxx solution.

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𝑦 = xxxxxxxxxxxxxxxxxxxxxxx 𝑥xxxx Y is the function. Now we xxxx to xxxx xxx xxxxxxxxxxx 𝑑𝑦

𝑑𝑥

𝑑𝑦

𝑑𝑥 x 𝑦′ x 1log8(log7(log6(log5 𝑥))) xxx xx 10 xxx xxxxxxxxxxxxxxxxxxxx 𝑥)))]′ x xxxxxxxxxxxxxxxxxxxx 𝑥xxx ∙ ln xx ∙ 1log7(log6(log5 𝑥)) xxx ln 8 xxxxxxxxxxxxxxx 𝑥xxx′ = 1log8(log7(log6(log5 𝑥xxx xxx ln 10 ∙ 1log7(log6(log5 𝑥)) ∙ ln x xxx xxxxxxxxxx 𝑥x xxx ln 7 [log6(log5 𝑥)]′ x 1log8(log7(log6(log5 𝑥xxx xxx xx xx xxx xxxxxxxxxxxxxxx 𝑥)) xxx ln x xxx xxxxxxxxxx 𝑥) ∙ ln 7

∙ 1log5 𝑥 ∙ xx x [log5 𝑥x′ x xxxxxxxxxxxxxxxxxxxx 𝑥))) ∙ ln 10 ∙ xxxxxxxxxxxxxxx 𝑥xx ∙ xx 8 ∙ xxxxxxxxxx 𝑥) ∙ ln x ∙ 1log5 𝑥 xxx ln xx

xxxxxx

𝑥 ∙ xx 5 xxx xx order to xxxx the domain xx xxxx xxxxxxxxxx we xxxx xxxxxxxx conditions. Firslty all xxxxxxxxx of xxxxxxxxxx xxxx xx greater xxxx xxxxx

(1) log7(log6(log5 𝑥xx > x 7log7(log6(log5 𝑥xx > xx log6(log5 𝑥) > 1 xxxxxxxxxx 𝑥x > xx log5 𝑥 > x 5log5 𝑥 > 56 𝑥 > xx

xxx log6(log5 𝑥x > 0 6log6(log5 𝑥) > 60 log5 𝑥 > 1 5log5 𝑥 > xx

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## domain, range and plot of the function in the attached file. All steps shown.

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All xxxxx xxxxxx

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Steps:

xxxxx

Domain:

xxxxxx

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## answer

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First, we should find xxxxx x is defined.

Note xxxx xxx domain of xxx xxx function is always xxx xxxxxxxx real xxxxxxx xxxxxxx xx are xxxxxxx with xxxxxxx logarithms xx which we are xxxxx

xxxxxxxxxxxxxxxxx>x

Log7log6log5x>xxxxxx

Log6log5x>xxxxxxxxx

xxxxx>6( 7)(8)(0)=6(7)

x>5(6)(7)(8)(0)=5(6)(7)

xxxxxx

Therefore any xxxxx xx x xxxx xxxx is xx the domain. Since xxx x xx a differentiable function, xxxxxxxxxxx is differentiable, and xx on. xxx xxxxxx xx xxxxx is xxx same xx xxx domain of xx

xxxxxxxx in bracket xxx raise to xxx xxxxx to xxx first

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