Find the derivative: frac{d}{dx} left[ log le | vedclass

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Question

(C) To find the derivative of $\log \left( x + \frac{1}{x} \right)$,we use the chain rule: $\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} \right) \r

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To find the derivative of $\log \left( x + \frac{1}{x} ight)$,we use the chain rule: $\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} ight) ight] = \frac{1}{x + \frac{1}{x}} 	imes \frac{d}{dx} \left( x + \frac{1}{x} ight)$ Since $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2}$,we have: $= \frac{1}{x + \frac{1}{x}} 	imes \left( 1 - \frac{1}{x^2} ight)$ $= \frac{1 - \frac{1}{x^2}}{x + \frac{1}{x}}$ Thus,the correct option is $C$.

Find the derivative: $\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} \right) \right] = $

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