If $f (x) = x cos x$ , find $f^{''} (x)$ .

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Solution

Using the product rule, we have
$f^{'} (x) = x \frac{d}{d x} (cos x) + cos x \frac{d}{d x} (x) = - x sin x + cos x$
To find $f^{''} (x)$ , we differentiate $f^{'} (x)$ :
$f^{''} (x) = \frac{d}{d x} (- x sin x + cos x)$
$= - x \frac{d}{d x} (sin x) + sin x \frac{d}{d x} (- x) + \frac{d}{d x} (cos x)$
$= - x cos x - sin x - sin x = - x cos x - 2 sin x$ .

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If f(x)=xcosx, find f′′(x).

Using the product rule, we havef′(x)=xddx(cosx)+cosxddx(x)=−xsinx+cosxTo find f′′(x), we differentiate f′(x):f′′(x)=ddx(−xsinx+cosx)=−xddx(sinx)+sinxddx(−x)+ddx(cosx)=−xcosx−sinx−sinx=−xcosx−2sinx.

If $f (x) = x cos x$ , find $f^{''} (x)$ .