To avoid having to do first the quotient rule and then the product rule on the bottom, I will take advantage of the fact that #cot(x)=1/tan(x)# and rewrite the function like this:
#tan(x)/(cos^2(x)cot(x))=tan(x)/(cos^2(x)/tan(x))=tan(x)*tan(x)/cos^2(x)=tan^2(x)/cos^2(x)#
The quotient rule states that:
#d/dx(f(x)/g(x))=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#
In our case, #f(x)=tan^2(x)# and #g(x)=cos^2(x)#. The derivatives are:
#f'(x)=2tan(x)sec^2(x)# (by the chain rule)
#g'(x)=-2cos(x)sin(x)# (again, by the chain rule)
Our derivative then becomes:
#(2tan(x)sec^2(x)cos^2(x)-(-2sin(x)cos(x)tan^2(x)))/(cos^4(x))#
To simplify this, I will use thee two identities:
#tan(theta)=sin(theta)/cos(theta)#
#sec(theta)=1/cos(theta)#
#(2tan(x)1/cancel(cos^2(x))cancel(cos^2(x))+2sin(x)cancel(cos(x))sin^2(x)/cos^cancel(2)(x))/cos^4(x)#
#(2tan(x)+2sin^3(x)/cos(x))/cos^4(x)=(2tan(x)+2sin^2(x)tan(x))/cos^4(x)#
