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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?

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    $\begingroup$ Your just deriving the quotient rule on the fly, rather than assuming it is true and then applying it directly; there is nothing wrong with doing the former. $\endgroup$
    – user785085
    May 8, 2021 at 13:39
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    $\begingroup$ First thing a mathematician would do is write $$y=1+\frac{x^2+2}{2x^3+3x^2}$$ before taking the derivative. $\endgroup$ May 8, 2021 at 16:14
  • $\begingroup$ What do you mean "I think both answers are correct."? If you do the math properly, you'll get the same answer with both methods. $\endgroup$ May 9, 2021 at 0:10
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    $\begingroup$ Real physicists and mathematicians stick the whole thing into their computer algebra system and don't worry about the precise algorithms it follows, unless they have a particular reason to risk silly errors by doing that kind of calculation by hand. Symbolic differentiation is a solved problem; one doesn't earn any "purity points" in the real world by doing it the hard way. $\endgroup$ May 9, 2021 at 0:33
  • $\begingroup$ @JosephSible-ReinstateMonica Yes! I got same answer :) $\endgroup$
    – user876873
    May 9, 2021 at 4:30

3 Answers 3

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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

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  • $\begingroup$ That's very helpful. But, it was my main question 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? $\endgroup$
    – user876873
    May 8, 2021 at 13:44
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    $\begingroup$ @Istiak Please don't keep making that bold. Sure, it's OK, because any valid proof is OK; there are no "rules" beyond that. To be honest, I hardly ever see mathematicians or physicists explicitly using the quotient rule. It seems to be more for teaching than anything else. $\endgroup$
    – J.G.
    May 8, 2021 at 13:45
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    $\begingroup$ I'd always use the quotient rule on a quotient because it is usually much simpler to work with $g'$ than $(\frac1g)'$ $\endgroup$ May 8, 2021 at 13:45
  • $\begingroup$ @J.G. Actually, I had read in a meta post that is saying,you should bold when it may need to be attracted or, something just like this. I don't remember. That's why I was just making that bold $\endgroup$
    – user876873
    May 8, 2021 at 13:48
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    $\begingroup$ @Istiak I see, cool. Thanks, though, for your edit that means you now only bold it in the question. $\endgroup$
    – J.G.
    May 8, 2021 at 13:52
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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.

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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 $$\left (\frac 1 h \right)'=-h' \cdot \frac 1 {h^2} = -\frac {h'} {h^2}.$$

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    $\begingroup$ $g^{-1}$ is a poor choice of notation here, as it usually means the inverse of $g$, and not the reciprocal. $\endgroup$
    – Xander Henderson
    May 9, 2021 at 0:19
  • $\begingroup$ @XanderHenderson I fixed the notation, thanks... as you remember I never used the $g^{-1}$ notation in the followings $\endgroup$ May 6 at 15:12

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