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Question Number 191796 by mathlove last updated on 30/Apr/23

Commented by mathlove last updated on 01/May/23

$p l e a s s o l v e t h i s$
	Answered by deleteduser1 last updated on 01/May/23

	$A) y^{^{'}} = \frac{x^{2}}{2 \sqrt{x^{2} + 1}} + \frac{\sqrt{x^{2} + 1}}{2} + x + B$ $B = \frac{d}{d x} [I n \sqrt{x + \sqrt{x^{2} + 1}}] = \frac{1}{\sqrt{x + \sqrt{x^{2} + 1}}} [\frac{1}{2 \sqrt{x + \sqrt{x^{2} + 1}}} (1 + \frac{x}{\sqrt{x^{2} + 1}})]$ $= \frac{1}{2 (x + \sqrt{x^{2} + 1})} + \frac{x}{2 \sqrt{x^{2} + 1} (x + \sqrt{x^{2} + 1})}$ $\Rightarrow B = \frac{x + \sqrt{x^{2} + 1}}{2 \sqrt{x^{2} + 1} (x + \sqrt{x^{2} + 1})} = \frac{1}{2 \sqrt{x^{2} + 1}}$ $\Rightarrow y^{^{'}} = \frac{x^{2} + (x^{2} + 1) + 2 x \sqrt{x^{2} + 1} + 1}{2 \sqrt{x^{2} + 1}}$ $y^{'} = \frac{(x^{2} + x^{2} + 1 + 2 x \sqrt{x^{2} + 1} + 1)}{2 \sqrt{x^{2} + 1}}$ $y^{'} = \frac{x^{2} + x \sqrt{x^{2} + 1} + 1}{\sqrt{x^{2} + 1}} = \frac{(x^{2} + 1) (\sqrt{x^{2} + 1} + x)}{(x^{2} + 1)}$ $\Rightarrow y^{^{'}} = \sqrt{x^{2} + 1} + x$ $B) 2 y = x \sqrt{x^{2} + 1} + x^{2} + 2 I n \sqrt{x + \sqrt{x^{2} + 1}}$ ${x y}^{^{'}} + {I n y}^{^{'}} = x \sqrt{x^{2} + 1} + x^{2} + I n (\sqrt{x^{2} + 1} + x)$ $= x \sqrt{x^{2} + 1} + x^{2} + \frac{2 I n (\sqrt{x^{2} + 1} + x)}{2}$ $\Rightarrow {x y}^{^{'}} + {I n y}^{'} = x \sqrt{x^{2} + 1} + x^{2} + 2 I n \sqrt{x + \sqrt{x^{2} + 1}} = 2 y$
	Commented by mathlove last updated on 01/May/23

	$t h a n k s$
	Answered by manxsol last updated on 01/May/23

	$y = \frac{x}{2} (\sqrt{x^{2} + 1} + x) + \frac{1}{2} l n (\sqrt{x^{2} + 1} + x$ $y = \frac{1}{2} (x z + l n z) z = \sqrt{x^{2} + 1} + x$ $z^{'} = 1 + \frac{x}{\sqrt{x^{2} + 1}} = \frac{z}{\sqrt{x^{2} + 1}}$ $z^{2} = 2 x^{2} + 2 x \sqrt{x^{2} + 1} + 1$ $y^{'} = \frac{1}{2} (z + {x z}^{'} + \frac{z^{'}}{z})$ $y^{'} = \frac{1}{2} (z + x \frac{z}{\sqrt{x^{2} + 1}} + \frac{1}{\sqrt{x^{2} + 1}})$ $y^{'} = \frac{1}{2} (\frac{z^{2}}{\sqrt{x^{2} + 1}} + \frac{1}{\sqrt{x^{2} + 1}})$ $y^{'} = \frac{1}{2} (\frac{2 x^{2} + 2 x \sqrt{x^{2} + 1} + 2}{\sqrt{x^{2} + 1}})$ $y^{'} = \frac{x^{2} + 1}{\sqrt{x^{2} + 1}} + x$ $y^{'} = \sqrt{x^{2} + 1} + x = z$ $b)$ $y = \frac{1}{2} (x z + l n z)$ $2 y = {x y}^{'} + {l n y}^{'}$
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