Question

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Answer 1

Question:

$If \: y = \sqrt{tanx + \sqrt{tanx + \sqrt{tanx + ... \infty } } } \\ then \: prove \: that \: \frac{dy}{dx} = \frac{ {sec}^{2}x }{2y - 1} .$

Solution:

Please refer to the attachment.

Note:

1. Sum rule and difference rule:

If y = f(x) ± g(x) , then ;

dy/dx = d{f(x)}/dx ± d{g(x)}/dx

2. Product rule:

If y = f(x)•g(x) , then ;

dy/dx = g(x)•d{f(x)}/dx + f(x)•d{g(x)}/dx

3. Division rule:

If y = f(x)/g(x) , then ;

dy/dx =

[g(x)•d{f(x)}/dx - f(x)•d{g(x)}/dx] / {g(x)}²

4. Chain rule:

If y = f(u) , u = g(v) , v = h(x) , then ;

dy/dx = (dy/du)•(du/dv)•(dv/dx)