How do you differentiate #f(x)=sin(cos(tanx))# using the chain rule?

1 Answer
Dec 20, 2015

Answer:

#f'(x)=-sec^2xsin(tanx)cos(cos(tanx))#

Explanation:

The first major issue is the sin function. According to the chain rule,

#d/dx[sinu]=u'cosu#, so

#f'(x)=cos(cos(tanx))*d/dx[cos(tanx)]#

Now to differentiate #cos(tanx)#, know that

#d/dx[cosu]=-u'sinu#

Thus,

#d/dx[cos(tanx)]=-d/dx[tanx]*sin(tanx)#

Also recall that #d/dx[tanx]=sec^2x#, so

#d/dx[cos(tanx)]=-sec^2xsin(tanx)#

Plug this back into the #f'(x)# equation.

#f'(x)=-sec^2xsin(tanx)cos(cos(tanx))#