0
$\begingroup$

Lets say we have $f'(x)$ when $f(x)=(x^2+3)(x^3-1)$.

We could use product rule with $u = (x^2+3)$ and $v = (x^3-1)$, but we would get the same answer if we had just multiplied $uv$ before taking the derivative.

Does this apply to any problem where we take the derivative of two factors being multiplied and why? If so, why would we ever need to use the product rule? Does the same apply for the quotient rule (although i understand that for quotient rule, it would be easier to simplify when not dividing variables)?

$\endgroup$
  • 3
    $\begingroup$ you are thinking about polynomials only. product rule is useful to find a derivative of, for example, $f(x)=x^2e^x$ $\endgroup$ – Vasya Oct 9 '18 at 14:02
  • $\begingroup$ Do you want to get $f'((x^2+3)(x^3-1))$ or $f'(x)$ when $f(x)=(x^2+3)(x^3-1)$? $\endgroup$ – mfl Oct 9 '18 at 14:02
  • $\begingroup$ $2\cdot3=6$. This means that you can replace $2\cdot3$ by $6$, or conversely, in any correct formula. $\endgroup$ – Christian Blatter Oct 9 '18 at 15:51
5
$\begingroup$

Sure, you are always free to make any valid algebraic simplification at any time, e.g. expanding a product of polynomials. So a problem like this one can be done either by using the product rule, or by first multiplying out the polynomials and then using just the power and sum rules. Both methods will yield the same answer.

If so, why would we ever need to use the product rule?

Well, think about an example like

$$\frac{d}{dx} \left[(8x^6+3x^5-2x^4+x^3-9x^2+2x+3)(2x^5+x^4-x^3+4x^2-8x+4)\right]$$

You can certainly multiply out all 42 terms if you like, but I think you would find it much more convenient to use the product rule here.

Also, not all functions are polynomials; if you want to find $\frac{d}{dx} x \sin(x)$ there is not really an obvious way to "multiply it out".

Does the same apply for the quotient rule (although i understand that for quotient rule, it would be easier to simplify when not dividing variables)?

Sure, if the quotient can be simplified. But this is not always possible; if you look at $\frac{d}{dx} \frac{x^2+2x+2}{x^2+7x+12}$, the numerator and denominator have no common factors, so you can't really write it in a simpler way.

$\endgroup$
2
$\begingroup$

Yes, you can multiply them together first. $$f(x) = (x^2+3)(x^3-1) \implies f(x) = x^5-x^2+3x^3-3$$ $$f’x = 5x^4+9x^2-2x$$ Using the Product Rule gets the same answer. For questions regarding polynomials, using the it might not be necessary since multiplying them directly gives the answer quickly.

However, not all functions are in this form. As said in one of the comments, imagine you have a function like $f(x) = x^2e^x$.

How would you differentiate that function? Here, the Product Rule comes in handy. $$(fg)’ = f’g+g’f$$ Our $f(x)$ is $x^2$ and our $g(x)$ is $e^x$. We know that $f’x = 2x$ and $g’x = e^x$.

So, we can find the derivative of $x^2e^x$ using it.

$$(x^2e^x)’ = (x^2)’e^x+(e^x)’x^2$$ $$(x^2e^x)’ = 2xe^x+x^2e^x \implies \boxed{(x^2e^x)’ = e^x(2x+x^2)}$$

Or, for example, $f(x) = x\sin(x)$. Once again, we can’t multiply the terms, so we’ll use the rule again.

$$(x\sin(x))’ = x’\sin(x)+\sin’(x)x \implies \boxed{(x\sin(x))’ = \sin(x)+x\cos(x)}$$

As you can see, neither of those functions were polynomials. For such problems in which simplification via multiplication is impossible/hard, you can use the Product Rule.

The exact same idea applies to the Quotient Rule, which is essentially a “variation” of the Product Rule.

$\endgroup$
  • $\begingroup$ $\cos(x)\sin(x)$ might not be the best example since we can “multiply terms” to get $\frac12\sin(2x)$. $\endgroup$ – amd Oct 9 '18 at 18:34
  • $\begingroup$ Ah yes, didn’t think about possible simplification there. Thanks for pointing out. $\endgroup$ – KM101 Oct 9 '18 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.