difference between quotient rule and product rule Product rule :
$$\frac{d}{dx} \big(f(x)\cdot g(x)\big)=f'(x)\cdot g(x)+f(x)\cdot g' (x)$$
Quotient rule :
$$\frac{d}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$$
Suppose, the following is given in question.
$$y=\frac{2x^3+4x^2+2}{3x^2+2x^3}$$
Simply, this is looking like Quotient rule. But, if I follow arrange the equation following way
$$y=(2x^3+4x^2+2)(3x^2+2x^3)^{-1}$$
Then, we can solve it using Product rule. As I was solving earlier problems in a pdf book using Product rule. I think both answers are correct. But, my question is, How does a Physicist and Mathematician solve this type question? Even, is it OK to use Product rule instead of Quotient rule in University and Real Life?
 A: Note that $((g(x))^{-1})'=-g'(x)(g(x))^{-2}$. Then, applying the product rule: $$\left(\frac{f(x)}{g(x)}\right)'=\left(f(x)\cdot \frac{1}{g(x)}\right)'=\frac{f'(x)}{g(x)}+\frac{-g'(x)f(x)}{(g(x))^2}=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$$
which is the quotient rule
A: The answer to

How does a Physicist and Mathematician solve this type question? Even,
is it OK to use Product rule instead of Quotient rule in University
and Real Life?

is that any experienced scientist knows several methods to solve problems and uses those that are most convenient for them at that particular time.
I would look at that derivative and use the quotient rule. But if there was something in the source of the problem that suggested that it made more sense to write the denominator as $(3x^2+2x^3)^{-1}$ then the product rule would be more appropriate.
A: the quotient rule is not a separate and non-compatible rule. It is just the product rule inserting $\frac 1 g$ instead of $g$.
let's see...
to make it clear... I show you, prove it for you, how quotient rule is just compatible with product law.
$$\left(\frac f g\right)'=\left(f\cdot \frac 1 g\right)'$$
as product law says:
$$=f' \cdot \frac 1 g +f\cdot \left (\frac 1 g\right)'$$
while $\displaystyle\left(\frac 1 g \right)' = \frac {-g'}{g^2}$, we would have
$$=f'\cdot \frac 1 g+f\cdot \frac{-g'}{g^2}=\frac{f'\cdot g-f\cdot g'}{g^2}.$$

Note:
You may know that $\displaystyle\left(\frac 1 h \right)' = \frac {-h'}{h^2}$ could be calculated by product rule, as if one consider the product $\displaystyle\left(\frac 1 h \cdot h \right) = 1$, and calculate the derivative of both sides of the equation. one the left hand side we have a constant which may already know the derivative is $0$, but on the other side we see a product so by applying the product rule we have
$$0=h' \cdot \frac 1 h + h\cdot \left (\frac 1 h \right)'.$$
Therefore
$$h\cdot \left (\frac 1 h \right)'=-h' \cdot \frac 1 h.$$
And so
$$\left (\frac 1 h \right)'=-h' \cdot \frac 1 {h^2} = -\frac {h'} {h^2}.$$
