If u=(fog)(x) and v=(gof)(x), verify if u'=v'? f(x)=3x+2,g(x)=x^2+1
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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'.
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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.
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