Find Devivaties with respect to x:

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SUMMARY

The discussion focuses on finding derivatives with respect to the variable x for two functions: y=sin(cos(x+3)) and tan(x+y)=ln x+5y. The first derivative for the first function is correctly calculated as y' = -sin(x+3)cos(cos(x+3)). For the second function, the derivative was initially miscalculated; the correct expression is y' = ((1/x) - Sec^2(x+y))/(Sec^2(x+y) - 5y), with the error identified in the treatment of the term (5y)'. The participants confirm the correctness of the first problem and provide constructive feedback on the second.

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

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