Differentiate the following functions w.r. t.x. f (x) = sq.root of x square + 1

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Solution

$f (x) = \sqrt{x^{2} + 1} Differentiate both side w . r . t . x, we get \frac{d}{dx} f (x) = \frac{d}{dx} \sqrt{x^{2} + 1} \Rightarrow f' (x) = \frac{d}{dx} {(x^{2} + 1)}^{\frac{1}{2}} = \frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1} \frac{d}{dx} (x^{2} + 1) = \frac{1}{2} {(x^{2} + 1)}^{- \frac{1}{2}} (2 x + 0) = \frac{1}{2 \sqrt{x^{2} + 1}} (2 x) = \frac{x}{\sqrt{x^{2} + 1}} Hence, f' (x) = \frac{x}{\sqrt{x^{2} + 1}}$

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