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Chapter 6: Problem 16

If \(y=\left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right)\), prove that \(2 x y \frac{d y}{d x}=\frac{x}{a}-\frac{a}{x}\)

Short Answer

Expert verified
The derivative of \(y\) with respect to \(x\) is \(-\frac{1}{2 \sqrt{a x}} \cdot (\frac{x}{a} + \frac{a}{x})\). Thus \(2xy \frac{d y}{d x}\) is indeed \(\frac{x}{a} - \frac{a}{x}\), proving the given equality.

Step by step solution

01

- Simplification of \( y \) before differentiation

Rewrite \( y \) as \( y = \sqrt{\frac{x^2}{a x}} + \sqrt{\frac{a^2}{a x}} \) = \( \frac{x}{\sqrt{a x}} + \frac{a}{\sqrt{a x}} \), to separate out \( x \) and \( a \) in the denominators.
02

- Differentiation

Now find \( \frac{d y}{d x} \). By applying the quotient and chain rules of differentiation, \( \frac{d y}{d x} = \frac{1}{2} \left( -\frac{x}{2(a x)^\frac{3}{2}} - \frac{a}{2(a x)^\frac{3}{2}} \right) = -\frac{1}{2 \sqrt{a x}} \left( \frac{x}{a x} + \frac{a}{a x} \right) = -\frac{1}{2 \sqrt{a x}} \cdot (\frac{x}{a} + \frac{a}{x}) \).
03

- Calculation

Calculate \(2xy \frac{d y}{d x}\) using \(y = \frac{x}{\sqrt{a x}} + \frac{a}{\sqrt{a x}}\) and \(\frac{d y}{d x}\) from previous step : \(2 x (\frac{x}{\sqrt{a x}} + \frac{a}{\sqrt{a x}}) \cdot (-\frac{1}{2 \sqrt{a x}} \cdot (\frac{x}{a} + \frac{a}{x})) = - x \cdot (\frac{x}{\sqrt{a x}} + \frac{a}{\sqrt{a x}}) \cdot (\frac{x}{a} + \frac{a}{x}) = - x \cdot \( \frac{1}{\sqrt{a x}} \left( \frac{x^2}{a} + a \right) = -x \cdot (\frac{x}{a} + \frac{a}{x}) = \frac{-x^2}{a} - \frac{a x}{x} = \frac{x}{a} - \frac{a}{x}\)

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