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Mathematics
OpenStudy (anonymous):

This is a very simple question

OpenStudy (anonymous):

Where is the question?

OpenStudy (anonymous):

\[\sqrt{sinx +\sqrt{sinx+\sqrt{sinx+....\to \infty}}}\] what is its derivative

OpenStudy (anonymous):

Mani I know you can solve this sum in a sec so just give the answer not the steps

OpenStudy (mani_jha):

cosx, I guess

OpenStudy (anonymous):

Mani I didn't expect this from you

OpenStudy (mani_jha):

Oh wait, cosx/(2S-1) where S is the above function

OpenStudy (anonymous):

Why did you fail in the first try

OpenStudy (mani_jha):

I did the calculation in my mind; because you asked not to do the steps. So, I made a mistake :\

OpenStudy (anonymous):

Cheers!!!!!

OpenStudy (anonymous):

\[y = \sqrt{sinx +\sqrt{sinx + ...}}\] 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.

OpenStudy (anonymous):

love these types of problems...

OpenStudy (anonymous):

I know they look huge but actually the are so little its almost too easy

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