Question

urgent. cat este f(x)= ln x+1 supra x-1 derivata. f'(x) cu cat este egal? va rog. explicit. si demonstrati ca functia este convexa pe intervalul(1,+infinit) va rooog. multumesc!

Answer 1

[tex]\\F_{(x)} = ln\frac{x+1}{x-1} \\F`_{(x)} = [\frac{ln(x+1)}{x-1}]` \\Aplicam \:\ regula \:\ lui \:\ Chain: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \\ f=\ln \left(u\right),\:\:u=\frac{x+1}{x-1} \\ Acum \:\ derivam ln(u) si u=\frac{x+1}{x-1} \\ ln(u)` = \boxed{\frac{1}{u}} \\ iar \:\ u` = \frac{\frac{}{}\left(x+1\right)`\left(x-1\right)-\frac{}{}\left(x-1\right)\left(x+1\right)`}{\left(x-1\right)^2} = \boxed{-\frac{2}{\left(x-1\right)^2}}[/tex]

[tex]\\Acum \:\ vine: \frac{1}{u}\left(-\frac{2}{\left(x-1\right)^2}\right) = \frac{1}{\frac{x+1}{x-1}}\left(-\frac{2}{\left(x-1\right)^2}\right) = \\=\ \textgreater \ Aplicam \boxed{\left(-a\right)=-a} =\ \textgreater \ -\frac{1}{\frac{x+1}{x-1}}\cdot \frac{2}{\left(x-1\right)^2} \\=\ \textgreater \ Aplicam \boxed{\frac{1}{\frac{b}{c}}=\frac{c}{b}} =\ \textgreater \ =-\frac{x-1}{x+1}\cdot \frac{2}{\left(x-1\right)^2}} \\ =\ \textgreater \ \boxed{=-\frac{2}{\left(x+1\right)\left(x-1\right)}}[/tex]

[tex]\\ Demonstratie \:\ ca \:\ functia \:\ este \:\ convexa \\ \\ F``_{(x)}= [ -\frac{2}{\left(x+1\right)\left(x-1\right)} ]` =\ \textgreater \ \\ Dam \:\ constanta \:\ afara: \left(a\cdot f\right)'=a\cdot f' =\ \textgreater \ -2\left(\frac{1}{\left(x+1\right)\left(x-1\right)}\right)` \\ =\ \textgreater \ Aplicam \:\ regula \:\ exponentului \:\ \boxed{\frac{1}{a}=a^{-1}} \\ =\ \textgreater \ 2\left(\left(\left(x+1\right)\left(x-1\right)\right)^{-1}\right)` \\ =\ \textgreater \ Aplicam \:\ regula \:\ lui \:\ Chian \:\ iar: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}[/tex]



[tex]\\=\ \textgreater \ f=u^{-1},\:\:u=\left(x+1\right)\left(x-1\right) \\=\ \textgreater \ -2\frac{d}{du}\left(u^{-1}\right)\frac{d}{dx}\left(\left(x+1\right)\left(x-1\right)\right) \\=\ \textgreater \ Aplicam \:\ regula \:\ puterii : \quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \\ =\ \textgreater \ 1\cdot \:u^{-1-1} = -\frac{1}{u^2} \\ Acum =\ \textgreater \ -2\left(-\frac{1}{u^2}\right)\cdot \:2x \\ u=\left(x+1\right)\left(x-1\right) \\ -2\left(-\frac{1}{\left(\left(x+1\right)\left(x-1\right)\right)^2}\right)\cdot \:2x = \frac{4x}{\left(x+1\right)^2\left(x-1\right)^2}[/tex]

[tex]\\ =\ \textgreater \ \frac{4x}{\left(x+1\right)^2\left(x-1\right)^2} = \frac{4x}{x^4-2x^2+1} \\ \boxed{Am \:\ atasat \:\ tabelul !}[/tex]