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Show-that-the-curve-y-ln-5-7x-8-x-has-no-stationary-point-for-all-real-values-of-x-

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Question Number 8244 by lepan last updated on 04/Oct/16
Show that the curve y=ln(((5−7x)/(8+x))) has no stationary point for all real values of x.
$${Show}\:{that}\:{the}\:{curve}\:{y}={ln}\left(\frac{\mathrm{5}−\mathrm{7}{x}}{\mathrm{8}+{x}}\right)\:{has} \\ $$$${no}\:{stationary}\:{point}\:{for}\:{all}\:{real}\:{values} \\ $$$${of}\:{x}. \\ $$
Answered by 123456 last updated on 06/Oct/16
y=ln ((5−7x)/(8+x)) (dy/dx)=(d/dx)ln ((5−7x)/(8+x)) u=((5−7x)/(8+x)) =((dln u)/dx) =((dln u)/du)∙(du/dx) =(1/u)∙(du/dx) =((8+x)/(5−7x))∙(d/dx) ((5−7x)/(8+x)) { ((v=5−7x)),((w=8+x)) :} =((8+x)/(5−7x))∙(d/dx) (v/w) (d/dx) (u/v)=(d/dx)(u∙(1/v))=(du/dx)∙(1/v)+u∙(d/dx) (1/v)=(du/dx)∙(1/v)−u∙((dv/x)/v^2 )=((u′v−uv′)/v^2 ) =((8+x)/(5−7x))∙(((dv/dx)∙w−v∙(dw/dx))/w^2 ) =((8+x)/(5−7x))∙((((d(5−7x))/dx)∙(8+x)−(5−7x)∙((d(8+x))/dx))/((8+x)^2 )) =(1/(5−7x))∙((−7∙(8+x)−(5−7x)∙1)/(8+x)) =((−7∙8−7x−5+7x)/((5−7x)(8+x))) =−((61)/((5−7x)(8+x)))
$${y}=\mathrm{ln}\:\frac{\mathrm{5}−\mathrm{7}{x}}{\mathrm{8}+{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}}{{dx}}\mathrm{ln}\:\frac{\mathrm{5}−\mathrm{7}{x}}{\mathrm{8}+{x}}\:\:\:\:\:{u}=\frac{\mathrm{5}−\mathrm{7}{x}}{\mathrm{8}+{x}} \\ $$$$=\frac{{d}\mathrm{ln}\:{u}}{{dx}} \\ $$$$=\frac{{d}\mathrm{ln}\:{u}}{{du}}\centerdot\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{{u}}\centerdot\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{8}+{x}}{\mathrm{5}−\mathrm{7}{x}}\centerdot\frac{{d}}{{dx}}\:\frac{\mathrm{5}−\mathrm{7}{x}}{\mathrm{8}+{x}}\:\:\:\:\:\:\:\:\:\begin{cases}{{v}=\mathrm{5}−\mathrm{7}{x}}\\{{w}=\mathrm{8}+{x}}\end{cases} \\ $$$$=\frac{\mathrm{8}+{x}}{\mathrm{5}−\mathrm{7}{x}}\centerdot\frac{{d}}{{dx}}\:\frac{{v}}{{w}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}}{{dx}}\:\frac{{u}}{{v}}=\frac{{d}}{{dx}}\left({u}\centerdot\frac{\mathrm{1}}{{v}}\right)=\frac{{du}}{{dx}}\centerdot\frac{\mathrm{1}}{{v}}+{u}\centerdot\frac{{d}}{{dx}}\:\frac{\mathrm{1}}{{v}}=\frac{{du}}{{dx}}\centerdot\frac{\mathrm{1}}{{v}}−{u}\centerdot\frac{{dv}/{x}}{{v}^{\mathrm{2}} }=\frac{{u}'{v}−{uv}'}{{v}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{8}+{x}}{\mathrm{5}−\mathrm{7}{x}}\centerdot\frac{\frac{{dv}}{{dx}}\centerdot{w}−{v}\centerdot\frac{{dw}}{{dx}}}{{w}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{8}+{x}}{\mathrm{5}−\mathrm{7}{x}}\centerdot\frac{\frac{{d}\left(\mathrm{5}−\mathrm{7}{x}\right)}{{dx}}\centerdot\left(\mathrm{8}+{x}\right)−\left(\mathrm{5}−\mathrm{7}{x}\right)\centerdot\frac{{d}\left(\mathrm{8}+{x}\right)}{{dx}}}{\left(\mathrm{8}+{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}−\mathrm{7}{x}}\centerdot\frac{−\mathrm{7}\centerdot\left(\mathrm{8}+{x}\right)−\left(\mathrm{5}−\mathrm{7}{x}\right)\centerdot\mathrm{1}}{\mathrm{8}+{x}} \\ $$$$=\frac{−\mathrm{7}\centerdot\mathrm{8}−\mathrm{7}{x}−\mathrm{5}+\mathrm{7}{x}}{\left(\mathrm{5}−\mathrm{7}{x}\right)\left(\mathrm{8}+{x}\right)} \\ $$$$=−\frac{\mathrm{61}}{\left(\mathrm{5}−\mathrm{7}{x}\right)\left(\mathrm{8}+{x}\right)} \\ $$

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