Question

$\small\mathtt\red{if \: y = \frac { a x ^ { 3 } } { ( x - a ) ( x - b ) ( x - c ) } + \frac { b x } { ( x - b ) ( x - c ) } + \frac { c } { ( x - c ) } + 1 \: prove \: that \:\left. \begin{array} { l } { \frac { y ^ { \prime } } { y } = \frac { 1 } { x } [ \frac { a } { a - x } + \frac { b } { b - x } + \frac { a } { c - x } ] } \end{array} \right.}$
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Answer 1

Answer:

it's your answer!!!

hope it is correct!

Answer 2

Appropriate Question :-

$\sf \: If \: y = \frac { a x ^ { 3 } } { ( x - a ) ( x - b ) ( x - c ) } + \frac { b x } { ( x - b ) ( x - c ) } + \frac { c } { ( x - c ) } + 1, \: prove \: that$

$\rm :\longmapsto\:\dfrac{y'}{y} = \dfrac{1}{x} \bigg[ \dfrac{a}{(a - x)} + \dfrac{b}{(b - x)} + \dfrac{c}{(c - x)}\bigg]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } } { ( x - a ) ( x - b ) ( x - c ) } + \dfrac { b x } { ( x - b ) ( x - c ) } + \dfrac { c } { ( x - c ) } + 1$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } } { ( x - a ) ( x - b ) ( x - c ) } + \dfrac { b x } { ( x - b ) ( x - c ) } + \dfrac { c + x - c } { ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } } { ( x - a ) ( x - b ) ( x - c ) } + \dfrac { b x } { ( x - b ) ( x - c ) } + \dfrac {x} { ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } } { ( x - a ) ( x - b ) ( x - c ) } + \dfrac { b x + x(x - b)} { ( x - b ) ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } } { ( x - a ) ( x - b ) ( x - c ) } + \dfrac { b x + {x}^{2} - bx} { ( x - b ) ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } } { ( x - a ) ( x - b ) ( x - c ) } + \dfrac {{x}^{2}} { ( x - b ) ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } + {x}^{2}(x - a) } { ( x - a ) ( x - b ) ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac { a x ^ { 2 } + {x}^{3} - a {x}^{2} } { ( x - a ) ( x - b ) ( x - c ) }$

$\rm :\longmapsto\:\sf \: y = \dfrac {{x}^{3} } { ( x - a ) ( x - b ) ( x - c ) }$

On taking log both sides, we get

$\rm :\longmapsto\:\sf \: logy =log\bigg[ \dfrac {{x}^{3} } { ( x - a ) ( x - b ) ( x - c ) }\bigg]$

We know,

$\boxed{\tt{ log \frac{x}{y} = logx \: - \: logy \: }}$

and

$\boxed{\tt{ log {x}^{y} = y \: logx \: }}$

So, using this, we get

$\rm :\longmapsto\:logy = 3logx - log(x - a) - log(x - b) - log(x - c)$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}logy =\dfrac{d}{dx}\bigg[ 3logx - log(x - a) - log(x - b) - log(x - c)\bigg]$

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{3}{x} - \dfrac{1}{x - a} - \dfrac{1}{x - b} - \dfrac{1}{x - c}$

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{3}{x} + \dfrac{1}{a - x} + \dfrac{1}{b - x} + \dfrac{1}{c - x}$

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{1}{x} + \dfrac{1}{x} + \dfrac{1}{x} + \dfrac{1}{a - x} + \dfrac{1}{b - x} + \dfrac{1}{c - x}$

can be re-arranged as

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{1}{x} + \dfrac{1}{a - x} + \dfrac{1}{x} + \dfrac{1}{b - x} + \dfrac{1}{x} + \dfrac{1}{c - x}$

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{a - x + x}{x(a - x)} + \dfrac{x + b - x}{x(b - x)} + \dfrac{x - c + x}{x(c-x)}$

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{a}{x(a - x)} + \dfrac{b}{x(b - x)} + \dfrac{c}{x(c - x)}$

$\rm :\longmapsto\: \dfrac{1}{y}y' = \dfrac{1}{x} \bigg[ \dfrac{a}{(a - x)} + \dfrac{b}{(b - x)} + \dfrac{c}{(c - x)}\bigg]$

$\bf :\longmapsto\:\dfrac{y'}{y}= \dfrac{y}{x} \bigg[ \dfrac{a}{(a - x)} + \dfrac{b}{(b - x)} + \dfrac{c}{(c - x)}\bigg]$

$\green{\large{\boxed{\sf{Hence, Proved}}}}$

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More To Learn :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$