When we find functions of functions, the answer is the chain rule, 99% of the times. Let's recall it: if #h# and #k# are two differentiable functions on #]a,b[#, then #forall x in ]a,b[#

#[h(k(x))]^'=h^'(k(x))k^'(x)#

In this specific case

#h(u)=tan(u)#

#k(x)=x+sec(x)#

and the derivatives are

#h^'(u)=1/cos^2(u)=sec^2(u)#

#k^'(x)=1+sin(x)/cos^2(x)=1+sin(x)/cos(x) 1/cos(x)=1+tan(x)sec(x)#

So putting all together:

#f^'(x)=[tan(x+sec(x))]^'=sec^2(x+sec(x))[1+tan(x)sec(x)]#

**Note:** To derive the tangent function and the secant function, we can recall their definitions:

#[tan(x)]^'=[sin(x)/cos(x)]^'= [cos(x)cos(x)-sin(x)[-sin(x)]]/cos^2(x)=(cos^2(x)+sin^2(x))/(cos^2(x))=1/(cos^2(x))=[1/cos(x)]^2=sec^2(x)#

#[sec(x)]^'=[1/cos(x)]^'=-(-sin(x))/cos^2(x)=sin(x)/cos(x)*1/cos(x)=tan(x)sec(x)#