Question

Question #ae606

Answer 1

See explanation...

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)`