Question

Find
dy
dx
il y = \sin x +
✓sinx + √sin x + √sinx
+ ...​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

$\boxed{\bf \: \dfrac{dy}{dx} = \dfrac{cosx}{2y - 1} \: }$

Step-by-step explanation:

Given that,

$\sf \: y = \sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ...} } }$

On squaring both sides, we get

$\sf \: {y}^{2} = sinx + \sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ...} } }$

$\sf \: {y}^{2} = sinx + y$

$\sf \: {y}^{2} - y = sinx$

On differentiating both sides w. r. t. x, we get

$\sf \: \dfrac{d}{dx}[ {y}^{2} - y] = \dfrac{d}{dx}sinx$

$\sf \: 2y\dfrac{dy}{dx} - \dfrac{dy}{dx} = cosx$

$\sf \: (2y - 1) \dfrac{dy}{dx} = cosx$

$\implies\sf \: \boxed{\bf \: \dfrac{dy}{dx} = \dfrac{cosx}{2y - 1} \: }$

$\rule{190pt}{2pt}$

Additional Information:

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$