If u= ({x-1}-{x+1}) and v={x+1}+{x-1} then d u/d v=.

Correct Answer is : −1v{-1}{{v}}v−1

Correct Answer is : −1v{-1}{{v}}v−1

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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 136 of 150

Marks: +1, -0

u=\log (\sqrt{x-1}-\sqrt{x+1})

v=\sqrt{x+1}+\sqrt{x-1}

\frac{d u}{d v}=\ldots

Solution:

Concept:

Use the chain rule and the fact that

\frac{du}{dv} = \frac{du/dx}{dv/dx}

Explanation:

Given

u = \log(\sqrt{x-1} - \sqrt{x+1})

Differentiate with respect to

x

\frac{du}{dx} = \frac{1}{\sqrt{x-1} - \sqrt{x+1}} \cdot \frac{d}{dx}(\sqrt{x-1} - \sqrt{x+1})

= \frac{1}{\sqrt{x-1} - \sqrt{x+1}} \left( \frac{1}{2\sqrt{x-1}} - \frac{1}{2\sqrt{x+1}} \right)

Simplify the bracket:

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

Cancel the common factor:

\frac{du}{dx} = \frac{-1}{2\sqrt{x^2-1}}

Now,

v = \sqrt{x+1} + \sqrt{x-1}

Differentiate with respect to

x

\frac{dv}{dx} = \frac{1}{2\sqrt{x+1}} + \frac{1}{2\sqrt{x-1}} = \frac{\sqrt{x-1} + \sqrt{x+1}}{2\sqrt{x^2-1}}

Divide the derivatives:

\frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{-1}{2\sqrt{x^2-1}} \cdot \frac{2\sqrt{x^2-1}}{\sqrt{x-1} + \sqrt{x+1}}

= \frac{-1}{\sqrt{x-1} + \sqrt{x+1}} = \frac{-1}{v}

Answer:

\frac{du}{dv} = \frac{-1}{v}

So, the correct option is D. $\frac{-1}{\mathrm{v}}$ .

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