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# Find y'(π) where y= 1/[(1÷x)+sinx]?

Find y'(π) where y= 1/[(1÷x)+sinx]

I would greatly appreciate any help with this problem. Thank you.

### 1 Answer

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- az_lenderLv 78 years agoFavorite Answer
It might suit me to write the y somewhat differently.

Multiply both numerator and denominator by x.

y = x / [1 + x sin(x) ]

I'm not sure that's easier, but I do hate fractions within fractions.

y ' = { (1 + x sin(x)) (1) - (x) [ sin(x) + x cos(x) ] } / [1 + x sin(x) ]^2

OK, now I plug in pi:

y '(pi) = { (1+0) (1) - (pi) (0-pi) } / [ 1 + 0 ] ^2

= ( 1 + pi^2) / 1^2

Appears to be just 1 + pi^2, or about 11

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