Let f(x)=3x^2+x+3/2x+1. What is the derivative of f with respect to x

If the function is f(x) = (3x+x+3)/(3x+1), that if f(x) = g(x)/h(x).
The derivative of this is f'(x) = [h(x)g'(x) - g(x)h'(x)]/h(x).

Since we have g(x) = 3x + x + 3, we can see that g'(x) = 6x + 1.
Since we have h(x) = 2x + 1, we can see that h'(x) = 2 and h(x) = (2x+1).

This makes the derivative be f'(x) = [(2x+1)(6x+1) - (3x+x+3)(2)]/(2x+1).
Multiplying the top out gives f'(x) = (12x+8x+1 - 3x-x-3)/(2x+1).
Combining terms gives f'(x) = (9x+7x-2)/(2x+1).

That factors out to f'(x) = (9x-2)(x+1)/(2x+1).
As can be seen, nothing cancels.


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