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

1 Answer
Dec 16, 2015

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

Explanation:

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#