1. Sep 9, 2012 #1

   tommy2st

   

   Find Devivaties with respect to x:
   
   1). y=sin(cos(x+3));
   
   2). tan(x+y)=ln x+ 5y.
   
   heres my work...just need a double check
   
   1). y=sin(cos(x+3));
   
   2). tan(x+y)=ln x+ 5y.
   
   1:y = sin(cos(x+3))
   y' = cos(cos(x+3))(-sin(x+3))(1)
   y' = {-sin(x+3)}{cos(cos(x+3))}
   
   2. Tan(x+y) = lnx+5y
   Sec^2(x+y)(1+y') = 1/x + 5yy'
   y' = {((1/x))-(Sec^2(x+y))}/(Sec^2(x+y) - 5y)

    

   tommy2st, Sep 9, 2012
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3. Sep 9, 2012 #2

   Klungo

   

   You messed up in deriving 5y on problem 2. (5y)' = 5y'. Not 5yy'. Everything else prior is correct and so is problem 1.

    

   Klungo, Sep 9, 2012

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