We have the functions f(x) = 3x+ 2 and g(x) = x^2 + 1

u = fog ( x) = f(g(x))

=> f(x^2 + 1)

=> 3(x^2 + 1) + 2

=> 3x^2 + 3 + 2

=> 3x^2 + 5

v = gof(x) = g(f(x))

=> g( 3x + 2)

=> (3x + 2)^2 +1

=> 9x^2 + 4 + 12x + 1

=> 9x^2 + 12x + 5

u' = 6x

v' = 18x + 12

**Therefore u' is not equal to v'.**

The answer is very simple: the derivatives of the functions u and v are not equal.

Since we know that the composition of 2 functions is not commutative, then the derivatives of 2 expressions that are not matching, are not equal.

We'll see how it happens.

We'll compose (fog)(x) = f(g(x))

We'll substitute x by g(x), in the expression of f(x).

f(g(x)) = 3(x^2 + 1) + 2

We'll remove the brackets:

f(g(x)) = 3x^2 + 5

u(x) = f(g(x))

We'll differentiate with respect to x:

u'(x) = 6x

Now, we'll compose (gof)(x) = g(f(x))

We'll substitute x by f(x), in the expression of g(x).

g(f(x)) = (3x + 2)^2 + 1

We'll expand the square:

g(f(x)) = 9x^2 + 12x + 4 + 1

g(f(x)) = 9x^2 + 12x + 5

v(x) = g(f(x))

We'll differentiate with respect to x:

v'(x) = 18x + 12

**We can see that the expression of u'(x) = 6x and the expression of v'(x) = 18x + 12 are not equal.**