How do you differentiate $f(x) =cos x-xtanx$ ?

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Answer 1 · 2015-12-16T00:42:13

$f'(x)=-sinx-tanx-xsec^2x$

Find the derivative of each term, then find their sum, which is equal to $f'(x)$ .

$d/dx[cosx]=-sinx$

The following derivative requires the chain rule.

$d/dx[xtanx]=tanxd/dx[x]+xd/dx[tanx]$

$d/dx[x]=1$
$d/dx[tanx]=sec^2x$

Thus,

$d/dx[xtanx]=tanx+xsec^2x$

Add the two derivatives to find that

$f'(x)=-sinx-tanx-xsec^2x$

How do you differentiate f(x) =cos x-xtanx f(x) =cos x-xtanx ?