Question

(i) If $y=\frac{a{x}^2}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}+\frac{bx}{\left(x-b\right)\left(x-c\right)}+\frac{c}{x-c}+1$, 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,
$y=\frac{a{x}^2}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}+\frac{bx}{\left(x-b\right)\left(x-c\right)}+\frac{c}{x-c}+1$
$\Rightarrow y=\frac{a{x}^2+bx\left(x-a\right)+c\left(x-a\right)\left(x-b\right)+\left(x-a\right)\left(x-b\right)\left(x-c\right)}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}$
$\Rightarrow y=\frac{x^3}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}$
$\Rightarrow logy=log\left\{\frac{x^3}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}\right\}$
$\Rightarrow logy=3logx-\left\{log\left(x-a\right)+log\left(x-b\right)+log\left(x-c\right)\right\}$
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 $\frac{dy}{dx}=2\frac{d}{dx}\left(\frac{x}{\sqrt{x^2-a^2}}\right)$
$=2\left\{\frac{\sqrt{x^2-a^2}\times 1-x\cdot \frac{1}{2}{\left(x^2-a^2\right)}^{-1/2}\frac{d}{dx}\left(x^2-a^2\right)}{x^2-a^2}\right\}$
$=2\left\{\frac{\sqrt{x^2-a^2}-\frac{x}{2\sqrt{x^2-a^2}}\left(2x-0\right)}{x^2-a^2}\right\}$
$=2\left\{\frac{x^2-a^2-x^2}{\sqrt{x^2-a^2\left(x^2-a^2\right)}}\right\}$
$=\frac{-2a^2}{{\left(x^2-a^2\right)}^{3/2}}$