Q.If y = 1 + a/(x-a) + bx/((x-a)(x-b)) + cx²/((x-a)(x-b)(x-c)), then show that dy/dx = (y/x)[a/(a-x) + b/(b-x) + c/(c-x)]. OR If (x-a)² + (y-b)² = c², then show that {1+(dy/dx)²}^(3/2) / (d²y/dx²) = -c.

The messy sum for $y$ telescopes into a clean closed form $y=x^3/[(x-a)(x-b)(x-c)]$ if you combine the terms two at a time; from there, logarithmic differentiation gives the required result directly.

Step 1 — collapse $y$ into closed form. Combine the first two terms:

$1+\frac{a}{x-a}=\frac{(x-a)+a}{x-a}=\frac{x}{x-a}.$

Add the third term:

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

Add the fourth term:

$\frac{x^2}{(x-a)(x-b)}+\frac{cx^2}{(x-a)(x-b)(x-c)}=\frac{x^2(x-c)+cx^2}{(x-a)(x-b)(x-c)}=\frac{x^3-cx^2+cx^2}{(x-a)(x-b)(x-c)}=\frac{x^3}{(x-a)(x-b)(x-c)}.$

So $y=\dfrac{x^3}{(x-a)(x-b)(x-c)}$.

Step 2 — logarithmic differentiation. Since $y$ is a product/quotient of powers, take $\log$ of both sides:

$\ln y = 3\ln x-\ln(x-a)-\ln(x-b)-\ln(x-c).$

Differentiate with respect to $x$:

$\frac{1}{y}\frac{dy}{dx}=\frac3x-\frac{1}{x-a}-\frac{1}{x-b}-\frac{1}{x-c}=\frac3x+\frac{1}{a-x}+\frac{1}{b-x}+\frac{1}{c-x}.$

Step 3 — match the required form. Note $\dfrac{a}{a-x}=\dfrac{(a-x)+x}{a-x}=1+\dfrac{x}{a-x}$, so

$\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}=3+x\left(\frac{1}{a-x}+\frac{1}{b-x}+\frac{1}{c-x}\right).$ …