Question #ae606

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Question #ae606

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Answer 1 · 2017-08-29T02:45:53

See explanation...

Explanation:

remember the chain rule:

$\frac{d}{d x} f (g (x)) = \frac{d f}{d g} \cdot \frac{d g}{d x}$ $d/dx f(g(x)) = (df)/(dg)*(dg)/dx$

so, if $f (u) = e^{u}$ $f(u) = e^u$ and $u = x cos x$ $u = xcosx$ ,

then $\frac{d f}{d u} = e^{u}$ $(df)/(du) = e^u$ .

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

$\frac{d}{d x} (f (x) g (x)) = \frac{d f}{d x} g (x) + \frac{d g}{d x} f (x)$ $d/(dx) (f(x)g(x)) = (df)/dx g(x) + (dg)/dx f(x)$
so, $\frac{d}{d x} x cos x = cos x - x sin x$ $d/(dx) xcosx = cosx -xsinx$

$\frac{d}{d x} e^{x cos x} = e^{x cos x} (cos x - x sin x)$ $d/dx e^(xcosx)= e^(xcosx) ( cosx -xsinx)$

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Question #ae606

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