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

$S h o w t h a t t h e c u r v e y = l n (\frac{5 - 7 x}{8 + x}) h a s$ $n o s t a t i o n a r y p o i n t f o r a l l r e a l v a l u e s$ $o f x .$
	Answered by 123456 last updated on 06/Oct/16
	$y=ln ((5−7x)/(8+x)) (dy/dx)=(d/dx)ln ((5−7x)/(8+x)) u=((5−7x)/(8+x)) =((dln u)/dx) =((dln u)/du)∙(du/dx) =(1/u)∙(du/dx) =((8+x)/(5−7x))∙(d/dx) ((5−7x)/(8+x)) { ((v=5−7x)),((w=8+x)) :} =((8+x)/(5−7x))∙(d/dx) (v/w) (d/dx) (u/v)=(d/dx)(u∙(1/v))=(du/dx)∙(1/v)+u∙(d/dx) (1/v)=(du/dx)∙(1/v)−u∙((dv/x)/v^2 )=((u′v−uv′)/v^2 ) =((8+x)/(5−7x))∙(((dv/dx)∙w−v∙(dw/dx))/w^2 ) =((8+x)/(5−7x))∙((((d(5−7x))/dx)∙(8+x)−(5−7x)∙((d(8+x))/dx))/((8+x)^2 )) =(1/(5−7x))∙((−7∙(8+x)−(5−7x)∙1)/(8+x)) =((−7∙8−7x−5+7x)/((5−7x)(8+x))) =−((61)/((5−7x)(8+x)))$
	$y = \ln \frac{5 - 7 x}{8 + x}$ $\frac{d y}{d x} = \frac{d}{d x} \ln \frac{5 - 7 x}{8 + x} u = \frac{5 - 7 x}{8 + x}$ $= \frac{d \ln u}{d x}$ $= \frac{d \ln u}{d u} \cdot \frac{d u}{d x}$ $= \frac{1}{u} \cdot \frac{d u}{d x}$ $= \frac{8 + x}{5 - 7 x} \cdot \frac{d}{d x} \frac{5 - 7 x}{8 + x} {\begin{cases} v = 5 - 7 x \\ w = 8 + x \end{cases}$ $= \frac{8 + x}{5 - 7 x} \cdot \frac{d}{d x} \frac{v}{w} \frac{d}{d x} \frac{u}{v} = \frac{d}{d x} (u \cdot \frac{1}{v}) = \frac{d u}{d x} \cdot \frac{1}{v} + u \cdot \frac{d}{d x} \frac{1}{v} = \frac{d u}{d x} \cdot \frac{1}{v} - u \cdot \frac{d v / x}{v^{2}} = \frac{u^{'} v - {u v}^{'}}{v^{2}}$ $= \frac{8 + x}{5 - 7 x} \cdot \frac{\frac{d v}{d x} \cdot w - v \cdot \frac{d w}{d x}}{w^{2}}$ $= \frac{8 + x}{5 - 7 x} \cdot \frac{\frac{d (5 - 7 x)}{d x} \cdot (8 + x) - (5 - 7 x) \cdot \frac{d (8 + x)}{d x}}{{(8 + x)}^{2}}$ $= \frac{1}{5 - 7 x} \cdot \frac{- 7 \cdot (8 + x) - (5 - 7 x) \cdot 1}{8 + x}$ $= \frac{- 7 \cdot 8 - 7 x - 5 + 7 x}{(5 - 7 x) (8 + x)}$ $= - \frac{61}{(5 - 7 x) (8 + x)}$
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