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

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Answer 1 · 2015-12-20T15:45:01

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

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.

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

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