Ugly Differentiation 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 $$\frac{dy}{dx}=\frac{y}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$$
What is the smartest way to solve this?

HINT:
$$\dfrac c{x-c}+1=\dfrac x{x-c}$$
$$\dfrac{bx}{(x-b)(x-c)}+\dfrac x{x-c}=x\cdot\dfrac{b+(x-b)}{(x-b)(x-c)}=\dfrac{x^2}{(x-b)(x-c)}$$
$$\frac{ax^2}{(x-a)(x-b)(x-c)}+\dfrac{x^2}{(x-b)(x-c)}=x^2\cdot\frac{a+(x-a)}{(x-a)(x-b)(x-c)}=\dfrac{x^3}{(x-a)(x-b)(x-c)}$$
Now take logarithm in both sides and differentiate 
