y-x-x-x-y-

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 + \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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