Question

Q.
$if \: y = {(x + \frac{1}{x}) }^{1 + \frac{1}{x} } \: then \: find \: \frac{dy}{dx}$


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Answer 1

$y = ( {x + \frac{1}{x} })^{1 + \frac{1}{x} } \\ \\ we \: hav \: to \: find \: differentiation . \\ \\ we \: know.... \\ differentiation \: of \: {x}^{n} \: with \: respect \: \\ to \: x....\: \: \: \: {x}^{n} = \: n {x}^{n - 1} \\ \\ so....here.... \\ \\ \frac{dy}{dx} = (1 + \frac{1}{x} ) {(x + \frac{1}{x} )}^{1 + \frac{1}{x} - 1} \\ \\ \frac{dy}{dx} = ( \frac{x + 1}{x} ) {(x + \frac{1}{x}) }^{ \frac{1}{x} }$

Answer 2

Answer:

\begin{gathered}y = ( {x + \frac{1}{x} })^{1 + \frac{1}{x} } \\ \\ we \: hav \: to \: find \: differentiation . \\ \\ we \: know.... \\ differentiation \: of \: {x}^{n} \: with \: respect \: \\ to \: x....\: \: \: \: {x}^{n} = \: n {x}^{n - 1} \\ \\ so....here.... \\ \\ \frac{dy}{dx} = (1 + \frac{1}{x} ) {(x + \frac{1}{x} )}^{1 + \frac{1}{x} - 1} \\ \\ \frac{dy}{dx} = ( \frac{x + 1}{x} ) {(x + \frac{1}{x}) }^{ \frac{1}{x} } \\ \\ \end{gathered}