Question

If $y=\frac{{ax}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1$ then prove that $\frac{dy}{dx}=\frac{y}{x}[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}]$

Answer 1

$y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1$

$\Rightarrow y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\left(\frac{c+x-c}{x-c}\right)$

$\Rightarrow y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{x}{x-c}$

$\Rightarrow y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx+x(x-b)}{(x-b)(x-c)}$

$\Rightarrow y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{x^2}{(x-b)(x-c)}$

$\Rightarrow y=\frac{a{x}^2+x^2(x-a)}{(x-a)(x-b)(x-c)}$

$\Rightarrow y=\frac{x^3}{(x-a)(x-b)(x-c)}$

$\Rightarrow \log y=\log [\frac{x^3}{(x-a)(x-b)(x-c)}]$

$\Rightarrow \log y=3\log x-[\log (x-a)+\log (x-b)+\log (x-c)]$

Differentiate w.r.t $x$, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{3}{x}-[\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}]$

$\frac{dy}{dx}=y\left[\right(\frac{1}{x}-\frac{1}{x-a})+(\frac{1}{x}-\frac{1}{x-b})+(\frac{1}{x}-\frac{1}{x-c}\left)\right]$

$\frac{dy}{dx}=y[-\frac{a}{x(x-a)}-\frac{b}{x(x-b)}-\frac{c}{x(x-c)}]$

$\frac{dy}{dx}=\frac{y}{x}[-\frac{a}{x-a}-\frac{b}{x-b}-\frac{c}{x-c}]$

$\frac{dy}{dx}=\frac{y}{x}[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}]$