Question

(i) If $y=\frac{ax^{2}}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}+\frac{bx}{\left(x-b\right)\left(x-c\right)}+\frac{c}{x-c}+1$, then $\frac{dy}{dx}=$
(ii) Find $\frac{dy}{dx}$, when $y=\frac{\left(\sqrt{x+a}-\sqrt{x-a}\right)^{2}}{\sqrt{x^{2}-a^{2}}}$, where $x > a > 0$.

	(i)	(ii)
(a)	$\frac{y}{x}\left[\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\right]\,$	$\frac{2a}{\sqrt{x^{2}-a^{2}}}$
(b)	$\frac{x}{y}\left[\frac{a-x}{a^{2}-b}+\frac{b}{b-x}+\frac{c}{c-x}\right]\,$	$\frac{-2a^2}{\sqrt{x^{2}-a^{2}}}$
(c)	$\frac{y}{x}\left[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right]\,$	$\frac{-2a^{2}}{\left(x^{2}-a^{2}\right)^{3/2}}$
(d)	$\frac{y}{x}\left[\frac{a}{a-x}+\frac{b}{b-x}-\frac{c}{c-x}\right]\,$	$\frac{-2a^{2}}{\left(x^{2}-a^{2}\right)^{1/2}}$

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

Answer: (C)

(i) We have,
$y=\frac{ax^{2}}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}+\frac{bx}{\left(x-b\right)\left(x-c\right)}+\frac{c}{x-c}+1$
$\Rightarrow y=\frac{ax^{2}+bx\left(x-a\right)+c\left(x-a\right)\left(x-b\right)+\left(x-a\right)\left(x-b\right)\left(x-c\right)}{\left(x -a\right) \left(x -b\right) \left(x -c\right)}$
$\Rightarrow y=\frac{x^{3}}{\left(x -a\right) \left(x -b\right) \left(x -c\right)}$
$\Rightarrow log\,y = log\left\{\frac{x^{3}}{\left(x - a\right)\left(x -b\right) \left(x-c\right)}\right\}$
$\Rightarrow logy = 3logx - \left\{log\left(x - a\right) + log\left(x - b\right) + log\left(x - c\right)\right\}$
On differentiating w.r.t. to $x$, we get
$\frac{1}{y} \frac{dy}{dx}=\frac{3}{x}-\left\{\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}\right\}$
$\Rightarrow \frac{dy}{dx}=y\left\{\left(\frac{1}{x}-\frac{1}{x-a}\right)+\left(\frac{1}{x}-\frac{1}{x-b}\right)+\left(\frac{1}{x}-\frac{1}{x-c}\right)\right\}$
$\Rightarrow \frac{dy}{dx}=y\left\{\frac{-a}{x\left(x-a\right)}+-\frac{b}{x\left(x-b\right)}+\frac{-c}{x\left(x-c\right)}\right\}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}\left\{\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right\}$
(ii) Given $y=\frac{\left(\sqrt{x+a}-\sqrt{x-a}\right)^{2}}{\sqrt{x^{2}-a^{2}}}$
$\Rightarrow y=\frac{2x-2\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}-a^{2}}}=\frac{2x}{\sqrt{x^{2}-a^{2}}}-\frac{2\sqrt{x^{2}-a^{2}}}{\sqrt{x^{2}-a^{2}}}$
$\Rightarrow y=\frac{2x}{\sqrt{x^{2}-a^{2}}}-2$;
Differentiating w.r.t. $x$, we get $\frac{dy}{dx}=2 \frac{d}{dx}\left(\frac{x}{\sqrt{x^{2}-a^{2}}}\right)$
$=2\left\{\frac{\sqrt{x^{2}-a^{2}}\times1-x \cdot\frac{1}{2}\left(x^{2}-a^{2}\right)^{-1/2} \frac{d}{dx}\left(x^{2}-a^{2}\right)}{x^{2}-a^{2}}\right\}$
$=2\left\{\frac{\sqrt{x^{2}-a^{2}}-\frac{x}{2\sqrt{x^{2}-a^{2}}}\left(2x-0\right)}{x^{2}-a^{2}}\right\}$
$=2\left\{\frac{x^{2}-a^{2}-x^{2}}{\sqrt{x^{2}-a^{2}\left(x^{2}-a^{2}\right)}}\right\}$
$=\frac{-2a^{2}}{\left(x^{2}-a^{2}\right)^{3/2}}$