Question about determining whether to use the chain rule or not?

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The discussion centers on the differentiation of the function 6*sqrt(x^5) and the application of the chain rule. It is established that the chain rule is necessary when dealing with composite functions, as seen in the expression 6*sqrt(x^5), which can be represented as f(g(x)) where f(x)= 6*sqrt(x) and g(x)= x^5. Alternatively, rewriting the function as 6x^(5/2) allows the use of the power rule for differentiation, yielding the same derivative of 15x^(3/2). The key takeaway is that recognizing composite functions is crucial for determining the appropriate differentiation method.

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  • Familiarity with the power rule for differentiation
  • Basic algebraic manipulation of functions
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  • Study the application of the chain rule in various composite functions
  • Practice differentiating functions using the power rule
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For example,

if you differentiate 6*sqrt(x^5), would you use the chain rule? If not, why?

Thank you!
 
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JessicaJ283782 said:
For example,

if you differentiate 6*sqrt(x^5), would you use the chain rule? If not, why?

Thank you!
Should you use chain rule? It depends.

As your function is written, you have a composite function (a function whose argument is another function). To differentiate such a function requires the chain rule.

If you write your function as 6x5/2, though, now it's no longer a composite, so you could use the power rule (and also the constant multiple rule).
 
It's not a matter of applying some "hard and fast" rule. You use your knowledge and think about each individual problem. Any time you can see something that can be thought of as a composition of two (or more) functions, that is candidate for the chain rule. To differentiate [itex]6\sqrt{x^5}[/itex] you can think of its as f(g(x)) where [itex]f(x)= 6\sqrt{x}[/itex] and [itex]g(x)= x^5[/itex].

In that case, [itex]g'(x)= 5x^4[/itex] and [itex]f'(x)= (6x^{1/2})'= 6(1/2)x^{-1/2}= 3/\sqrt{x}[/itex] so the derivative is [itex](3/\sqrt{x^5})(5x^4)= 15(x^{-5/2})(x^4)[/itex][itex]= 15x^{-5/2+ 4}= 15x^{3/2}[/itex].


But, in this particular case, it is easier to do as Mark44 suggested: write the function as [itex]6(x^5)^{1/2}= 6x^{5/2}[/itex] so its derivative is [itex]6(5/2)x^{5/2- 1}= 15x^{3/2}[/itex] as above
 

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