y = (√(x+(√(x+(√x))))) y′ =


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Question Number 168276 by Florian last updated on 07/Apr/22

$y = \sqrt{x + \sqrt{x + \sqrt{x}}}$ $y^{'} =$
Commented by cortano1 last updated on 07/Apr/22

$y^{'} = \frac{1 + \frac{1 + \frac{1}{2 \sqrt{x}}}{2 \sqrt{x + \sqrt{x}}}}{2 \sqrt{x + \sqrt{x + \sqrt{x}}}}$
Commented by Florian last updated on 08/Apr/22

$? ? ?$
	Answered by greogoury55 last updated on 07/Apr/22

	$y = \sqrt{x + \sqrt{x + \sqrt{x}}}$ $y^{2} - x = \sqrt{x + \sqrt{x}}$ ${(y^{2} - x)}^{2} - x = \sqrt{x}$ $\frac{d}{d x} ({(y^{2} - x)}^{2} - x) = \frac{d}{d x} (\sqrt{x})$ $2 (y^{2} - x) (2 {y y}^{'} - 1) - 1 = \frac{1}{2 \sqrt{x}}$ $\Rightarrow 2 (y^{2} - x) (2 {y y}^{'} - 1) = \frac{1 + 2 \sqrt{x}}{2 \sqrt{x}}$ $\Rightarrow 2 {y y}^{'} - 1 = \frac{1 + 2 \sqrt{x}}{4 \sqrt{x} \sqrt{x + \sqrt{x}}} = \frac{1 + 2 \sqrt{x}}{4 \sqrt{x^{2} + x \sqrt{x}}}$ $\Rightarrow 2 {y y}^{'} = \frac{1 + 2 \sqrt{x} + 4 \sqrt{x^{2} + x \sqrt{x}}}{4 \sqrt{x^{2} + x \sqrt{x}}}$ $\Rightarrow y^{'} = \frac{1 + 2 \sqrt{x} + 4 \sqrt{x^{2} + x \sqrt{x}}}{8 \sqrt{x^{2} + x \sqrt{x}} \sqrt{x + \sqrt{x + \sqrt{x}}}}$ $\Rightarrow y^{'} = \frac{1 + 2 \sqrt{x} + 4 \sqrt{x^{2} + x \sqrt{x}}}{8 \sqrt{(x^{2} + x \sqrt{x}) (x + \sqrt{x + \sqrt{x}})}}$
	Commented by Florian last updated on 07/Apr/22

	$V e r y G o o d!$
	Answered by Mathspace last updated on 07/Apr/22
	$y^2 =x+(√(x+(√x))) ⇒(y^2 −x)^2 =x+(√x) ⇒ 2(2yy^′ −1)(y^2 −x)=1+(1/(2(√x))) ⇒ 4yy^′ =((2(√x)+1)/(2(√x)(y^2 −x)))=((2(√x)+1)/(2(√x)(√(x+(√x)))))+2 ⇒y^′ =(1/(4y)){((3(√x)+1)/(2(√(x^2 +x(√x))))) +2}$
	$y^{2} = x + \sqrt{x + \sqrt{x}}$ $\Rightarrow {(y^{2} - x)}^{2} = x + \sqrt{x} \Rightarrow$ $2 (2 {y y}^{^{'}} - 1) (y^{2} - x) = 1 + \frac{1}{2 \sqrt{x}} \Rightarrow$ $4 {y y}^{^{'}} = \frac{2 \sqrt{x} + 1}{2 \sqrt{x} (y^{2} - x)} = \frac{2 \sqrt{x} + 1}{2 \sqrt{x} \sqrt{x + \sqrt{x}}} + 2$ $\Rightarrow y^{^{'}} = \frac{1}{4 y} {\frac{3 \sqrt{x} + 1}{2 \sqrt{x^{2} + x \sqrt{x}}} + 2}$
	Answered by Florian last updated on 07/Apr/22

	$y = \sqrt{x + \sqrt{x + \sqrt{x}}}$ $y^{'} = {(\sqrt{x + \sqrt{x + \sqrt{x}}})}^{'} \to g_{1} = x + \sqrt{x + \sqrt{x}}$ $y^{'} = {(\sqrt{g_{1}})}^{'} {(x + \sqrt{x + \sqrt{x}})}^{'}$ $y^{'} = \frac{1}{2 \sqrt{g_{1}}} ({(x)}^{'} + {(\sqrt{x + \sqrt{x}})}^{'}) \to g_{2} = x + \sqrt{x}$ $y^{'} = \frac{1}{2 \sqrt{g_{1}}} (1 + {(\sqrt{g_{2}})}^{'} {(x + \sqrt{x})}^{'}$ $y^{'} = \frac{1}{2 \sqrt{g_{1}}} (1 + \frac{1}{2 \sqrt{g_{2}}} (1 + \frac{1}{2 \sqrt{x}})$ $y^{'} = \frac{1}{2 \sqrt{g_{1}}} (1 + \frac{1}{2 \sqrt{x + \sqrt{x}}} (1 + \frac{1}{2 \sqrt{x}}))$ $y^{'} = \frac{1}{2 \sqrt{g_{1}}} (1 + \frac{2 \sqrt{x + 1}}{4 \sqrt{x^{2} + \sqrt{x} x}})$ $y^{'} = \frac{1}{2 \sqrt{x + \sqrt{x + \sqrt{x}}}} (1 + \frac{2 \sqrt{x + 1}}{4 \sqrt{x^{2} + \sqrt{x} x}})$ $y^{'} = \frac{1}{2 \sqrt{x + \sqrt{x \sqrt{x}}}} \times \frac{4 \sqrt{x^{2} + \sqrt{x} x} + 2 \sqrt{x} + 1}{4 \sqrt{x^{2} + \sqrt{x} x}}$ $y^{'} = \frac{4 \sqrt{x^{2} + \sqrt{x} x} + 2 \sqrt{x} + 1}{8 \sqrt{(x + \sqrt{x + \sqrt{x}}) (x^{2} + \sqrt{x} x)}}$
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