The values of x,at which the first derivative | vedclass

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(A) Given the function $f(x) = (\sqrt{x} + \frac{1}{\sqrt{x}})^2$.Expanding the expression,we get $f(x) = x + \frac{1}{x} + 2$.Now,find the first deri

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$2, -2$

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(A) Given the function $f(x) = (\sqrt{x} + \frac{1}{\sqrt{x}})^2$.Expanding the expression,we get $f(x) = x + \frac{1}{x} + 2$.Now,find the first derivative $f'(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x + x^{-1} + 2) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.We are given that $f'(x) = \frac{3}{4}$.So,$1 - \frac{1}{x^2} = \frac{3}{4}$.Rearranging the terms,we get $\frac{1}{x^2} = 1 - \frac{3}{4} = \frac{1}{4}$.This implies $x^2 = 4$,which gives $x = \pm 2$.

The values of $x$,at which the first derivative of the function $(\sqrt{x} + \frac{1}{\sqrt{x}})^2$ with respect to $x$ is $\frac{3}{4}$,are

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