If y=≤ft[{x-a}+{x-b}] then \;d y/d x is

Correct Answer is : 12(x−a)(x−b)\;{1}{2{(x-a)(x-b)}}2(x−a)(x−b)1

Correct Answer is : 12(x−a)(x−b)\;{1}{2{(x-a)(x-b)}}2(x−a)(x−b)1

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Differential Calculus

Question : 21 of 69

Marks: +1, -0

y=\log\left[\sqrt{x-a}+\sqrt{x-b}\right]

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

Solution:

Concept:

Differentiate a logarithmic function using chain rule.

Explanation:

Let

y = \log\left[\sqrt{x-a} + \sqrt{x-b}\right]

Set

u = \sqrt{x-a} + \sqrt{x-b}

, so

y = \log u

Then

\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}

Compute

\frac{du}{dx} = \frac{1}{2\sqrt{x-a}} + \frac{1}{2\sqrt{x-b}} = \frac{1}{2}\left(\frac{1}{\sqrt{x-a}} + \frac{1}{\sqrt{x-b}}\right)

Combine the fractions:

\frac{1}{\sqrt{x-a}} + \frac{1}{\sqrt{x-b}} = \frac{\sqrt{x-b} + \sqrt{x-a}}{\sqrt{x-a}\sqrt{x-b}} = \frac{u}{\sqrt{(x-a)(x-b)}}

Thus

\frac{du}{dx} = \frac{1}{2} \cdot \frac{u}{\sqrt{(x-a)(x-b)}}

Therefore

\frac{dy}{dx} = \frac{1}{u} \cdot \frac{1}{2} \cdot \frac{u}{\sqrt{(x-a)(x-b)}} = \frac{1}{2\sqrt{(x-a)(x-b)}}

Answer:

\frac{1}{2\sqrt{(x-a)(x-b)}}

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