Question

plzz evaluate it as soon as possible ​

Answer 1

Answer:

Refer the attached picture.

Answer 2

Answer:

Step-by-step explanation:

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

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

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

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

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

$Hence$

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

Taking Logof both sides

$ln(y)=ln(x^{3} )-ln(x-a)-ln(x-b)-(ln([tex]</p><p>ln(y)=3ln(x )-ln(x-a)-ln(x-b)-(ln(x-c)$x-c)[/tex]

Differentiate with respect to x

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

Now write the term  $\frac{3}{x} as \frac{1}{x} +\frac{1}{x}+\frac{1}{x} ,$

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

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

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

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

And now multiple both the sides by y,

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