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I have a math problem where I am required to find the derivative of a function with the limitations of not being allowed to use the Product or Quotient Rule of Differentiation.

The problem looks like this:

$$h(x) = \frac{4-x^6}{3x^{-2}}$$

I have tried a variety of routes but always end up with results that seem to require the use of the Product or Quotient Rule.

For example, my latest try looks like this:

$$h(x) = \frac{4-x^6}{3x^{-2}}$$

$$h(x) = \frac{4}{3x^{-2}} - \frac{x^6}{3x^{-2}}$$

$$h(x) = \frac{4x^2}{3} - \frac{x^8}{3}$$

(From this step, I figured I could just use the Difference Rule of Differentiation, like this:)

$$h'(x) = \frac{d}{dx}\left(\frac{4x^2}{3}\right) - \frac{d}{dx}\left(\frac{x^8}{3}\right)$$

But wouldn't this actually end up using the Product -or- Quotient Rule? Like this:

$$h'(x) = \frac{d}{dx}\left(\frac{4}{3}(x^2)\right) - \frac{d}{dx}\left(\frac{1}{3}(x^8)\right)$$

Is there another route I can take with this type of problem that would avoid using the Product or Quotient Rule of Differentiation?

summea
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    "Product Rule" generally refers to finding the derivative of the product of two non-constant functions. I believe the approach you took is the one intended by whomever wrote the question. – Austin Mohr May 11 '13 at 04:47
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    You could alternately find the derivative from the definition of derivative. To save headaches, it is useful to make a preliminary simplification to $\frac{4}{3}x^2-\frac{1}{3}x^8$. You probably did the derivative of $x^2$ from the definition in class, the argument for the $\frac{4}{3}x^2$ part will be essentially the same. And the argument for the $\frac{1}{3}x^8$ part is not too bad, just expand $(x+h)^8$. – André Nicolas May 11 '13 at 05:12
  • https://math.stackexchange.com/a/4261705/945479 – qifeng618 Sep 28 '21 at 01:44

2 Answers2

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Most likely you are allowed to use the Constant Multiple Rule, i.e. $(cf(x))'=c(f(x))'$, where $c$ is a constant.

vadim123
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  • This looks like what the instructor intended; thank you for the direction! I can't believe I overlooked this rule... – summea May 11 '13 at 04:49
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Let $$h(x)=f(x)\cdot g(x)$$

So, $$\ln h(x)=\ln f(x)+\ln g(x)$$

Differentiating wrt $x$ using Chain Rule,

$$\frac{h'(x)}{h(x)}=\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}$$

$$\implies h'(x)=\frac{h(x)}{f(x)}\cdot f'(x)+\frac{h(x)}{g(x)}\cdot f'(x)=g(x)\cdot f'(x)+f(x)\cdot g'(x)$$

If $f(x)=c,$ (constant)

$f'(x)=0$ and $h(x)=c\cdot g(x)\implies h'(x)=c\cdot g'(x)$

lab bhattacharjee
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  • Eventually you are back to product rule . – Inceptio May 11 '13 at 04:52
  • @Inceptio That's what I was thinking... :/ I was intrigued about the use of ln(), though... – summea May 11 '13 at 04:52
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    Nice to mention logarithmic differentiation, a very useful idea. – André Nicolas May 11 '13 at 04:53
  • @Inceptio, put $f(x)=4-x^6, g(x)=\frac{x^2}3$ from the start – lab bhattacharjee May 11 '13 at 04:54
  • @labbhattacharjee I wish I could accept both answers on this question... but for the time being (and I'm sorry for not clarifying this in the original question,) I might accept the "Constant Multiple Rule" response because I haven't yet reached the Chain Rule in my particular class at school (although I know this is going to be a way to solve many things, in the future.) Thank you so much for your time and your response, though! – summea May 11 '13 at 05:00
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    @summea, sharing ideas & clearing confusions is much important than the acceptance of a solution. Btw, I've derived the constant multiplication formula, if you notice till last. – lab bhattacharjee May 11 '13 at 05:05
  • I'm a little late to the party, but doesn't this derivation of the product rule only work if $f$ and $g$ are strictly positive functions? – A. Howells Mar 19 '17 at 22:31