Find the value of Given, y = log(x + 1/x)Differentiate it with respect to x:dy/dx = [1/(x + 1/x)] ×d/dx(x + 1/x) 0ex0ex = [x/(x^2+1)] (1 – 1/x^2)0ex0e ...

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d/dx[logx + 1...

Question

$\frac{d}{d x} [\log (x + \frac{1}{x})] =$

$x + \frac{1}{x}$

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$\frac{(1 + \frac{1}{x^{2}})}{(x + \frac{1}{x})}$

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$\frac{(1 - \frac{1}{x^{2}})}{(x + \frac{1}{x})}$

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$1 + \frac{1}{x}$

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Solution

The correct option is C
$\frac{(1 - \frac{1}{x^{2}})}{(x + \frac{1}{x})}$

Find the value of $fraction numerator d over denominator d x end fraction space open square brackets log space open parentheses x space plus space 1 over x close parentheses close square brackets space equals$
Given, $y = \log (x + \frac{1}{x})$
Differentiate it with respect to $x$ :
$\frac{d y}{d x} = [\frac{1}{(x + \frac{1}{x})}] \times \frac{d}{d x} (x + \frac{1}{x}) = [\frac{x}{(x^{2} + 1)}] (1 - \frac{1}{x^{2}}) = [\frac{x}{(x^{2} + 1)}] \frac{(x^{2} - 1)}{x^{2}} = \frac{(x^{2} - 1)}{x (x^{2} + 1)} = \frac{x^{2} (1 - \frac{1}{x^{2}})}{x^{2} (x + \frac{1}{x})}$
$∴ \frac{d y}{d x} = \frac{(1 - \frac{1}{x^{2}})}{(x + \frac{1}{x})}$
Hence option (C) is correct.

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ddxlogx+1x=

The correct option is C (1-1x2)(x+1x)Find the value of Given, y=logx+1xDifferentiate it with respect to x:dydx=1x+1x×ddxx+1x=x(x2+1)1–1x2=x(x2+1)(x2–1)x2=(x2–1)x(x2+1)=x21–1x2x2x+1x∴dydx=(1–1x2)(x+1x)Hence option (C) is correct.

$\frac{d}{d x} [\log (x + \frac{1}{x})] =$