Logarithm / Exponential Calculus (1 Viewer)

atar90plus · Jun 16, 2012

Hello

Could you guys please help me with these questions relating to logarithm and exponential differentiation attached. Please show full working and solutions. Note: the questions I need help with is 18, 21, 23,24 and 27.

Thanks

Timske · Jun 16, 2012

Pictures are too blurry

<a href="http://www.codecogs.com/eqnedit.php?latex=21. ~(5 @plus; 2\ln x)^{4} - \textup{use ~chain ~rule} \\ 4(5@plus;2\ln x)^3*~\frac{2}{x} = \frac{8}{x}(5@plus;2\log x)^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?21. ~(5 + 2\ln x)^{4} - \textup{use ~chain ~rule} \\ 4(5+2\ln x)^3*~\frac{2}{x} = \frac{8}{x}(5+2\log x)^3" title="21. ~(5 + 2\ln x)^{4} - \textup{use ~chain ~rule} \\ 4(5+2\ln x)^3*~\frac{2}{x} = \frac{8}{x}(5+2\log x)^3" /></a>

Sy123 · Jun 16, 2012

Cant see the pictures man.

$20. \ \frac{d}{dx}\frac{10}{\ln{x}} \\ \\ = \frac{d}{dx}10(\ln{x})^{-1} \\ = 10 \times \frac{1}{x} \times (ln^{-2}{x}) \\ = \frac{10}{x\ln^2{x}}$

EDIT:

$23. \ \frac{d}{dx} \frac{2-3\ln{x}}{2+\ln{x}}.....[1] \\ \\ u=2-3\ln{x} \ \ \ u'=\frac{-3}{x} \\ v=2+3\ln{x} \ \ \ v'=\frac{3}{x} \\ \\ [1]=\frac{u'v-v'u}{v^2} \\ =\frac{\frac{-3}{x}(2+3\ln{x})-\frac{3}{x}(2-3\ln{x})}{(2+3\ln{x})^2}$

Logarithm / Exponential Calculus (1 Viewer)

atar90plus

01000101=YES! YES! YES!

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Timske

Sequential

Sy123

This too shall pass

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