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#