difference between quotient rule and product rule
Product rule :
$$\frac{d}{dx} f(x)g(x)=f'(x)g(x)+f(x)g' (x)$$
Quotient rule :
$$\frac{d}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$
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?
3 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
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.
0 comment threads
If you work out deriving the quotient rule yourself using the exact trick you're highlighting, you can see that the quotient rule is nothing more than the product rule and the chain rule used together. If you rearrange a quotient to use the product rule, then you'll very likely be using the chain rule shortly thereafter on the $(\cdots)^{-1}$ part, and you will inevitably arrive at exactly the same result as if you had used the quotient rule but with more steps. The only potential difference is whether your result looks like $\frac{a}{b} + \frac{c}{b^2}$ or $\frac{ab + c}{b^2}$—but of course, as I'm sure you can see, that's no difference at all.
0 comment threads
0 comment threads