Using sum difference derivative instead of prodict rule?

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Using the sum difference derivative instead of the product rule is valid for calculating the derivative of the function f(x)=(x^3-x)(x^2+2)(x^2+x-1). Multiplying the first two binomials and then applying the sum rule can simplify the process compared to using the product rule twice. The derivative can be expressed as (fgh)'= fgh'+ fg'h+ f'gh, which avoids unnecessary multiplication. Ultimately, both methods yield the same result, and the choice depends on which approach is simpler for the individual. It is important to ensure clarity and confidence in the chosen method.
Nano-Passion
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Let us say we have this:

f(x)=(x^3-x)(x^2+2)(x^2+x-1)

Would it be equally correct to multiply the first two binomials, and then taking that and multiplying by the last tri-nomial; and then using the sum rule? It seems perfectly fine (and much simpler!) but I want to make sure I am not violating anything here. It seems like a longer problem to use the product rule twice.
 
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Either way will work. Just guess what is most simple and start in.
 
Since you are talking about the derivative, I see no reason to multiply it out at all:
(fgh)'= fgh'+ fg'h+ f'gh.
((x^3−x)(x^2+2)(x^2+x−1))'= (x^3- x)(x^2+2)(2x+1)+ (x^3-x)(2x)(x^2+X-1)+ (3x-1)(x^2+2)(x^2+x-1)
 
HallsofIvy said:
Since you are talking about the derivative, I see no reason to multiply it out at all:
(fgh)'= fgh'+ fg'h+ f'gh.
((x^3−x)(x^2+2)(x^2+x−1))'= (x^3- x)(x^2+2)(2x+1)+ (x^3-x)(2x)(x^2+X-1)+ (3x-1)(x^2+2)(x^2+x-1)
True, I just figured I should simplify that more, so I wanted to do the other route. But now I see you don't have to simplify that more.

lurflurf said:
Either way will work. Just guess what is most simple and start in.

Okay, never hurts to be completely sure. :)
 

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