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Updated Question

Assuming I want to differentiate function using Chain Rule, $\frac {x^5}{(3+ 2x^{8})},$ The Chain Rule says, $(g\circ f)'(x) = f'(x)\cdot g'(f (x))$

So what's the logic or steps to determine $f(x)$ and $g(x)$?

PS: I have the answer using Quotient Rule.


Here is how I solve it finally using arbitrary function f(x) and g(x).

  1. separate $x^{5}$ as h(x)
  2. $f(x) = (3+2x^{8})$
  3. $g(x) = x^{-1} = g(f(x)) = (3+2x^{8})^{-1}$
  4. $\frac{d}{dx} h(x).g(x)$ = $\frac{d}{dx} x^{5}.[x^{-1}]$
  5. using Product rule, $\frac{d}{dx} x^{5}.[x^{-1}] = 5x^{4}.g(x) + x^{5}.g'(x)$; This g'(x) = the derivative of composition function, $(g\circ f)'(x)$

  6. applyg Chain rule to get g'(x), the composition function, $$(g\circ f)'(x)$$ = inside function's derivative . outside function's derivative.

  7. $f'(x) = 16x^{7}$

  8. $(g\circ f)'(x) = 16x^{7} . (-1)[x^{-2}]$
  9. plug in f(x) into outside function's x, $$(-1)[x^{-2}] = \frac{-1}{(3+2x^{8})^{2}} $$
  10. Thus, $(g\circ f)'(x) = 16x^{7} .\frac{-1}{(3+2x^{8})^{2}} $
  11. going back to where we pause at Product rule at Step 5 and applying each solved segment, $\frac{d}{dx} x^{5}.[x^{-1}] = 5x^{4}.g(x) + x^{5}.(g\circ f)'(x) = \frac{5x^{4}}{(3+2x^{8})} -\frac{16x^{7}}{(3+2x^{8})^{2}} $
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    $\begingroup$ The chain rule actually says $(g\circ f)'(x)=g'(f(x))f'(x)$. I don't know what you're doing... $\endgroup$
    – dfeuer
    Commented Oct 4, 2013 at 1:09
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    $\begingroup$ What you want to use is the quotient rule, really. $\endgroup$
    – Pedro
    Commented Oct 4, 2013 at 1:16
  • $\begingroup$ In the textbook it is solved using Chain Rule. I don't get jow it is done. Denominator has gone up as a -1 power. $\endgroup$
    – aspiring
    Commented Oct 4, 2013 at 2:01
  • $\begingroup$ This is much more easily solvable using quotient rule, but you can solve it using the chain rule by making up arbitrary $f(x)$ and $g(x)$ such that $f(g(x)) = \frac {x^5}{3 + 2x^8}$ $\endgroup$
    – MT_
    Commented Oct 4, 2013 at 2:17
  • $\begingroup$ DO NOT write $f'(g\circ f)$ if you mean $(g\circ f)'$. $\endgroup$ Commented Oct 4, 2013 at 3:53

1 Answer 1

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Your textbook is using the fact that $\dfrac{x^5}{3+2x^8}$ and $x^5(3+2x^8)^{-1}$ are two ways of writing the same thing.

Using the product and chain rule, \begin{align*} \frac{d}{dx}\left[x^5(3+2x^8)^{-1}\right] &= -x^5(3+2x^8)^{-2}(16x^7)+5x^4(3+2x^8)^{-1}\\ &=-16x^{12}(3+2x^8)^{-2}+5x^4(3+2x^8)^{-1} \\ &=x^4(3+2x^8)^{-2}\left[-16x^8+5(3+2x^8)\right]\\ &=3x^4(3+2x^8)^{-2}(5-2x^8)\\ &=\frac{3x^4(5-2x^8)}{(3+2x^8)^2} \end{align*}

Honestly, the quotient rule is more suited for this situation.

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    $\begingroup$ I suspect the original poster might have seen $\dfrac{x^5}{(3+2x)^8}$ and mis-copied it as $\dfrac{x^5}{(3+2x^8)}$. That would explain why the method used in the textbook relied on the chain rule. ${}\qquad{}$ $\endgroup$ Commented Oct 4, 2013 at 14:31
  • $\begingroup$ I assumed that since the OP wrote "the denominator went up as a -1 power," the expression was correctly copied. It is certainly strange however, most textbooks would simply use the quotient rule here. $\endgroup$
    – stochasm
    Commented Oct 4, 2013 at 14:57
  • $\begingroup$ @stochasm your assumption is indeed correct and appreciated for "decifering" my typo question. I followed the explaination you provided. First do the Product. Then do the inside, outside derivatives. $\endgroup$
    – aspiring
    Commented Oct 7, 2013 at 4:30
  • $\begingroup$ @stochasm I sort of had a doubt while using the steps above. At Step 5, with Prodcut rule I have $h'(x).g(x) + h(x).g'(x)$. However as I apply Chain Rule the $g'(x)$ is infact (g∘f)′(x). It's proven to be true in the computation. So is it correct to use that sort of notation reference? $\endgroup$
    – aspiring
    Commented Oct 7, 2013 at 5:09
  • $\begingroup$ There is no problem with using that kind of notation if it helps you find the derivatives of these types of functions more easily. However, you seem to have $g(x)$ defined as both $x^{-1}$ and $g(f(x))$, which makes what you wrote extremely difficult to understand. Choose a different name for $g(f(x))$ if you must. $\endgroup$
    – stochasm
    Commented Oct 7, 2013 at 10:26

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