Find dy/dx of y=Integral of a=1 to b=3x (tan^2t)dt

This problem is a substitution problem in Calculus 1 chapter 5.

I believe this is a fundamental theorem of calc problem
Were going to use $ as the integral
f=tan^2t
F=$f
Then y=F(b)-F(a)=F(3x)-F(1)

Then dy/dx= d/dx(F(3x))-d/dx(F(1))

Since F(1) is constant d/dx(F(1))=0

By the chain rule d/dx(F(3x))=3 f(3x)

So dy/dx=3tan^2(3x)

set g(x) = 3x then F(3x) = F(g(x)) then the chain rule applies d/dx(F(g(x))) = F'(g(x))*g'(x). F'(g(x)) = f(g(x)) and g'(x) = 3 (which should be obvious) then just substitute and you get the chain rule

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First let's set u=3x then
dy/dx=dy/du*du/dx=3*dy/du
and y=int(tan^2tdt) between t=1 and t=u.
dy/du=tan^2u using the fundamental theorem of calculus.
So dy/dx=3*tan^2u=3*tan^2(3x)