Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 2484 by John_Haha last updated on 21/Nov/15

find dy/dx of ln(sechx+lnlnx)

$${find}\:{dy}/{dx}\:{of}\:{ln}\left({sechx}+{lnlnx}\right) \\ $$

Answered by Filup last updated on 21/Nov/15

y=ln(sech(x)+ln(ln(x))) (dy/dx)=((d(ln(u)))/dx) (du/dx) (chain rule) u=sech(x)+ln(ln(x)) (du/dx)=(d/dx)((1/(cosh(x))))+(d/dx)(ln(ln(x))) (du/dx)=((d((1/k)))/dx) (1/k) (dk/dx)+(d/dx)(ln(ln(x))) k=cosh(x) =−sech(x)tanh(x)+((1/x)/(ln(x))) (du/dx)=(1/(xln(x)))−sech(x)tanh(x) (dy/dx)=(du/dx) (1/u) (dy/dx)=(((1/(xln(x)))−sech(x)tanh(x))/(sech(x)+ln(ln(x))))

$${y}=\mathrm{ln}\left(\mathrm{sech}\left({x}\right)+\mathrm{ln}\left(\mathrm{ln}\left({x}\right)\right)\right) \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}\left(\mathrm{ln}\left(\mathrm{u}\right)\right)}{\mathrm{d}{x}}\:\frac{{du}}{{dx}}\:\:\:\:\:\:\:\:\:\left({chain}\:{rule}\right) \\ $$$$ \\ $$$${u}=\mathrm{sech}\left({x}\right)+\mathrm{ln}\left(\mathrm{ln}\left({x}\right)\right) \\ $$$$\frac{{du}}{{dx}}=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{cosh}\left({x}\right)}\right)+\frac{{d}}{{dx}}\left(\mathrm{ln}\left(\mathrm{ln}\left({x}\right)\right)\right) \\ $$$$\frac{{du}}{{dx}}=\frac{{d}\left(\frac{\mathrm{1}}{{k}}\right)}{{dx}}\:\frac{\mathrm{1}}{{k}}\:\frac{{dk}}{{dx}}+\frac{{d}}{{dx}}\left(\mathrm{ln}\left(\mathrm{ln}\left({x}\right)\right)\right)\:\:\:\:\:\:{k}=\mathrm{cosh}\left({x}\right) \\ $$$$=−\mathrm{sech}\left({x}\right)\mathrm{tanh}\left({x}\right)+\frac{\frac{\mathrm{1}}{{x}}}{\mathrm{ln}\left({x}\right)} \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{1}}{{x}\mathrm{ln}\left({x}\right)}−\mathrm{sech}\left({x}\right)\mathrm{tanh}\left({x}\right) \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}\:\frac{\mathrm{1}}{{u}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\frac{\mathrm{1}}{{x}\mathrm{ln}\left({x}\right)}−\mathrm{sech}\left({x}\right)\mathrm{tanh}\left({x}\right)}{\mathrm{sech}\left({x}\right)+\mathrm{ln}\left(\mathrm{ln}\left({x}\right)\right)} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com