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

Relevance
  • 8 years ago
    Favorite 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

Still have questions? Get your answers by asking now.