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This is a very simple question
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Where is the question?
\[\sqrt{sinx +\sqrt{sinx+\sqrt{sinx+....\to \infty}}}\] what is its derivative
Mani I know you can solve this sum in a sec so just give the answer not the steps
cosx, I guess
Mani I didn't expect this from you
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Oh wait, cosx/(2S-1) where S is the above function
Why did you fail in the first try
I did the calculation in my mind; because you asked not to do the steps. So, I made a mistake :\
Cheers!!!!!
\[y = \sqrt{sinx +\sqrt{sinx + ...}}\] so \[y ^{2}=sinx+y\] differentiating implicitly gives 2yy' = cosx + y'. Solving for y' gives y'=cosx/(2y-1), where y is the function described above.
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love these types of problems...
I know they look huge but actually the are so little its almost too easy
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