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Question Number 167357 by mkam last updated on 13/Mar/22

find n^(th)  dervaiteve of ln(cosx) ?

$${find}\:{n}^{{th}} \:{dervaiteve}\:{of}\:{ln}\left({cosx}\right)\:? \\ $$

Answered by Coronavirus last updated on 16/Mar/22

   posons f(x)=ln(cos (x))    ∂f=−tan (x)  ∂^2 f  = −1−tan^2 (x)  ∂^3 f = −2tan(x) −2tan^3 (x)      ∂^4 f = −4tan^2 (x)−6tan^4 (x)+2   et omme tan(x)= sin(x)×(1/(cos(x) ))    apres je pense a^�  utiliser la flormule de Leibniz  ∂^n (f.g)=Σ_(i=0) ^n C_n ^i (∂^i f^ )(∂^( n−i) g)

$$\: \\ $$$${posons}\:{f}\left({x}\right)={ln}\left(\mathrm{cos}\:\left({x}\right)\right) \\ $$$$ \\ $$$$\partial{f}=−\mathrm{tan}\:\left({x}\right) \\ $$$$\partial^{\mathrm{2}} {f}\:\:=\:−\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\partial^{\mathrm{3}} {f}\:=\:−\mathrm{2tan}\left(\mathrm{x}\right)\:−\mathrm{2tan}^{\mathrm{3}} \left(\mathrm{x}\right)\:\:\:\: \\ $$$$\partial^{\mathrm{4}} {f}\:=\:−\mathrm{4tan}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{6tan}^{\mathrm{4}} \left(\mathrm{x}\right)+\mathrm{2}\: \\ $$$$\mathrm{et}\:\mathrm{omme}\:\mathrm{tan}\left(\mathrm{x}\right)=\:\mathrm{sin}\left(\mathrm{x}\right)×\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{x}\right)\:}\:\: \\ $$$${apres}\:{je}\:{pense}\:\grave {{a}}\:{utiliser}\:{la}\:{flormule}\:{de}\:{Leibniz} \\ $$$$\partial^{{n}} \left({f}.{g}\right)=\underset{{i}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{{n}} ^{{i}} \left(\partial^{{i}} {f}^{} \right)\left(\partial^{\:{n}−{i}} {g}\right) \\ $$

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