chain rule states that if f(x)=f(u(x)) then
\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}
as examples, first one
f(x)=\frac{7}{(4x^3-6x^2)^3}
u(x)=4x^3-6x^2
f(u)=\frac{7}{u^3}
given that
\frac{df}{du}=-\frac{21}{u^4}
and
\frac{du}{dx}=12x^2-12x
then chain rule states
\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=-\frac{21}{u^4}(12x^2-12x)
substituting u
\frac{df}{dx}=-\frac{252}{(4x^3-6x^2)^4}(x^2-x)
in the next one, you are almost there... all you have to do is factorize (6x^2-5)^4(2x-1)^3
the third one you can do it by calculating y=y(x) and x=x(y) and then differentiate ie
y(x)=\frac{x}{4}(3 \pm \sqrt{5})
or by implicit differentiation.
The next one you already have it, the only thing you need to do is evaluate y'(2) and y'(3) then use the fact that
m=\frac{f'(3)-f'(2)}{3-2}
in the last two you use the fact that (f\circ g)(x)=g(f(x)) be carefull tough, you are doing it wrong in the last one (watch the root).
