Math, asked by Shubhendu8898, 1 year ago

Find dy/dx
y = 1 + x/(x-a) + bx/(x-a)(x-b) + cx²/(x-a)(x-b)(x-c) .

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Answered by abhi569
42

y =  \huge{(} \small{1 + \dfrac{ a}{x - a} } \huge{)}  \small{    + \huge{(} \small{  \dfrac{ bx}{(x-a)(x-b)} + \dfrac{ cx^{2}}{( x - a)(x-b)(x-c)} }} \huge{)}




y =  \dfrac{x - a + a}{x - a}  +  \dfrac{bx(x - c) + cx {}^{2} }{(x - a)(x - b)(x - c)}  \\  \\  \\  \\  y =  \frac{x}{x - a}  +  \frac{b {x}^{2} - cbx + cx {}^{2}  }{(x - a)(x - b)(x - c)}  \\  \\  \\  \\  =  > y =  \frac{x(x - b)(x - c) + bx {}^{2}  - cbx + cx {}^{2} }{ (x - a)(x - b)(x - c)}



 =  >y =  \dfrac{ {x}^{3} -  {x}^{2}  c - bx {}^{2} + bcx   + bx {}^{2}  -    bcx + cx {}^{2} }{(x - a)(x - b)(x - c)}  \\  \\  \\  \\  =  >  y =  \frac{ {x}^{3} }{(x - a)(x - b)(x - c)}





Taking log on both sides,


 =  > log \: y = log \:  \dfrac{ {x}^{3} }{(x - a)(x - b)(x - c)}


From the properties of logarithms ,



  =  >  log \: y = log {x}^{3}  - log(x - a)    - log(x - b) - log(x - c)



From the properties of logarithms ,



  =  >  log \: y = 3log {x}^{}  - log(x - a)    - log(x - b) - log(x - c)



Then,

 =  >  \dfrac{ y_{1}}{y} =  \dfrac{3}{x}  +  \dfrac{1}{x - a}  +  \dfrac{1}{x - b}  +  \dfrac{1}{x - c}  \\  \\  \\  \\  =  >   \frac{dy}{dx}  = y \times ( \dfrac{ 1 }{ x }  -   \frac{1}{x - a}  +  \frac{1}{x}  -  \frac{1}{x - b}    +  \frac{1}{x}   - \frac{1}{x - c} ) \\
 \\  \\  \\  =  >  \frac{dy}{dx}  = y \times ( \frac{ x - a  - x}{x(x - a)}  +  \frac{x - b  -  x}{x(x  - b)   }  +  \frac{x - c  - x }{x(x - c)}  \\  \\  \\  \\   =  >  \frac{dy}{dx}  =  \frac{y}{x} \times  ( \frac{ - a}{x - a}  +  \frac{ - b}{x - b}  +  \frac{ - c}{x - c} ) \\  \\  \\  \\  =  >  \frac{dy}{dx}  =  \frac{y}{x}  \times ( \frac{a}{a - x}  +  \frac{b}{b - x}  +  \frac{c}{c - x} )






Please check it.



siddhartharao77: nice answer!
abhi569: :-)
QGP: Thanks for the answer :)
abhi569: :-)
Answered by Anonymous
1

Step-by-step explanation:

Then,

\begin{lgathered} \sf \: = &gt; \dfrac{ y_{1}}{y} = \dfrac{3}{x} + \dfrac{1}{x - a} + \dfrac{1}{x - b} + \dfrac{1}{x - c} \\ \\ \\ \\ \sf \:  = &gt; \frac{dy}{dx} = y \times ( \dfrac{ 1 }{ x } - \frac{1}{x - a} + \frac{1}{x} - \frac{1}{x - b} + \frac{1}{x} - \frac{1}{x - c} ) \\\end{lgathered} </p><p>

\begin{lgathered}\\ \\ \\  \sf \: = &gt; \frac{dy}{dx} = y \times ( \frac{ x - a - x}{x(x - a)} + \frac{x - b - x}{x(x - b) } + \frac{x - c - x }{x(x - c)} \\ \\ \\ \\  \sf \: = &gt; \frac{dy}{dx} = \frac{y}{x} \times ( \frac{ - a}{x - a} + \frac{ - b}{x - b} + \frac{ - c}{x - c} ) \\ \\ \\ \\ = &gt; \sf \red{\frac{dy}{dx} = \frac{y}{x} \times ( \frac{a}{a - x} + \frac{b}{b - x} + \frac{c}{c - x} )}\end{lgathered}

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