if y = \; {{x²+1} + {x²-1}}{{x²+1} - {x²-1}}, then \; d y/d x is

Correct Answer is : 2x+ 2x3x4−12x + \; {2x³}{{x^{4}-1}}2x+x4−12x3

Correct Answer is : 2x+ 2x3x4−12x + \; {2x³}{{x^{4}-1}}2x+x4−12x3

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Differential Calculus

Question : 32 of 69

Marks: +1, -0

y = \; \frac{\sqrt{x^{2}+1} + \sqrt{x^{2}-1}}{\sqrt{x^{2}+1} - \sqrt{x^{2}-1}},

\; \frac{d y}{d x}

Solution:

Concept:

Simplify the given expression by rationalizing the denominator to make differentiation easier.

Explanation:

First, rationalize the denominator:

y = \frac{\sqrt{x^{2}+1} + \sqrt{x^{2}-1}}{\sqrt{x^{2}+1} - \sqrt{x^{2}-1}} \times \frac{\sqrt{x^{2}+1} + \sqrt{x^{2}-1}}{\sqrt{x^{2}+1} + \sqrt{x^{2}-1}}

The denominator becomes

(\sqrt{x^{2}+1})^{2} - (\sqrt{x^{2}-1})^{2} = (x^{2}+1)-(x^{2}-1)=2

The numerator becomes

(\sqrt{x^{2}+1}+\sqrt{x^{2}-1})^{2} = (x^{2}+1)+(x^{2}-1)+2\sqrt{(x^{2}+1)(x^{2}-1)} = 2x^{2}+2\sqrt{x^{4}-1}

Thus

y = \frac{2x^{2}+2\sqrt{x^{4}-1}}{2} = x^{2} + \sqrt{x^{4}-1}

Now differentiate with respect to

x

\frac{dy}{dx} = 2x + \frac{1}{2\sqrt{x^{4}-1}} \cdot 4x^{3} = 2x + \frac{2x^{3}}{\sqrt{x^{4}-1}}

Answer:

2x + \frac{2x^{3}}{\sqrt{x^{4}-1}}

, which corresponds to option B.

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