How to determine inside and outside functions for Chain Rule?

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, $\frac{d}{dx} x^{5}.[x^{-1}] = 5x^{4}.g(x) + x^{5}.(g\circ f)'(x) = \frac{5x^{4}}{(3+2x^{8})} -\frac{16x^{7}}{(3+2x^{8})^{2}}$
• The chain rule actually says $(g\circ f)'(x)=g'(f(x))f'(x)$. I don't know what you're doing... – dfeuer Oct 4 '13 at 1:09
• What you want to use is the quotient rule, really. – Pedro Tamaroff Oct 4 '13 at 1:16
• 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. – aspiring Oct 4 '13 at 2:01
• 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 $f(g(x)) = \frac {x^5}{3 + 2x^8}$ – MCT Oct 4 '13 at 2:17
• DO NOT write $f'(g\circ f)$ if you mean $(g\circ f)'$. – Michael Hardy Oct 4 '13 at 3:53

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.
• 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. ${}\qquad{}$ – Michael Hardy Oct 4 '13 at 14:31
• @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? – aspiring Oct 7 '13 at 5:09
• 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. – stochasm Oct 7 '13 at 10:26