Question

Differentiate

\frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}

with respect to

x

.

Accepted Answer

The given function is $\frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$ .

Let $y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$

$y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}} \times \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}}$

$y = \frac{{(\sqrt{a + x} - \sqrt{a - x})}^{2}}{{(\sqrt{a + x})}^{2} - {(\sqrt{a - x})}^{2}}$

$y = \frac{a + x + a - x - 2 \sqrt{a^{2} - x^{2}}}{a + x - (a - x)} [∵ {(a - b)}^{2} = a^{2} + b^{2} - 2 a b]$

$y = \frac{2 a - 2 \sqrt{a^{2} - x^{2}}}{a + x - a + x}$

$y = \frac{2 a - 2 \sqrt{a^{2} - x^{2}}}{2 x}$

$y = \frac{a - \sqrt{a^{2} - x^{2}}}{x}$

On differentiating both sides with respect to $x$ , we get

$\frac{d}{d x} (y) = \frac{d}{d x} (\frac{a - \sqrt{a^{2} - x^{2}}}{x})$

$\frac{d y}{d x} = \frac{(x) \frac{d}{d x} (a - \sqrt{a^{2} - x^{2}}) - (a - \sqrt{a^{2} - x^{2}}) \frac{d}{d x} (x)}{{(x)}^{2}}$ [Using quotient rule]

$\frac{d y}{d x} = \frac{(x) [\frac{d}{d x} (a) - \frac{d}{d x} \sqrt{a^{2} - x^{2}}] - (a - \sqrt{a^{2} - x^{2}}) \cdot (1)}{{(x)}^{2}}$

$\frac{d y}{d x} = \frac{(x) [0 - \frac{1}{2 \sqrt{a^{2} - x^{2}}} \cdot \frac{d}{d x} (a^{2} - x^{2})] - (a - \sqrt{a^{2} - x^{2}})}{{(x)}^{2}}$

$\frac{d y}{d x} = \frac{(x) [- \frac{1}{2 \sqrt{a^{2} - x^{2}}} \cdot (- 2 x)] - (a - \sqrt{a^{2} - x^{2}})}{{(x)}^{2}} [∵ \frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}}; \frac{d}{d x} x^{n} = n x^{n - 1}]$

$\frac{d y}{d x} = [\frac{x^{2}}{\sqrt{a^{2} - x^{2}}} - (a - \sqrt{a^{2} - x^{2}})] \frac{1}{x^{2}}$

$\frac{d y}{d x} = \frac{x^{2} - a \sqrt{a^{2} - x^{2}} + a^{2} - x^{2}}{x^{2} \sqrt{a^{2} - x^{2}}}$

$\frac{d y}{d x} = \frac{a (a - \sqrt{a^{2} - x^{2}})}{x^{2} \sqrt{a^{2} - x^{2}}}$

M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 6: EXERCISE 5.6

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 6: EXERCISE 5.6

Questions from M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 6: EXERCISE 5.6 with Hints & Solutions

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