Question #ae606

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Answer 1 · 2017-08-29T02:45:53

See explanation...

remember the chain rule:

$d/dx f(g(x)) = (df)/(dg)*(dg)/dx$

so, if $f(u) = e^u$ and $u = xcosx$ ,

then $(df)/(du) = e^u$ .

To calculate (du)/(dx), we must use the rule for finding the derivative of the product of 2 functions.

$d/(dx) (f(x)g(x)) = (df)/dx g(x) + (dg)/dx f(x)$
so, $d/(dx) xcosx = cosx -xsinx$

$d/dx e^(xcosx)= e^(xcosx) ( cosx -xsinx)$