Why Is the Derivative of sin(ax²) Equal to 2axcos(ax²)?

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Discussion Overview

The discussion centers on understanding the derivative of the function sin(ax²), specifically addressing the components of the derivative, particularly the term 2ax. The scope includes mathematical reasoning and clarification of the chain rule in calculus.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • One participant asserts that the derivative of sin(ax²) is 2axcos(ax²) and expresses confusion about the origin of the 2ax term.
  • Another participant prompts the first to consider the chain rule as a method for finding the derivative.
  • A subsequent reply explains the application of the chain rule, identifying f(x) as sin(x) and g(x) as ax², and provides the formula for the derivative.
  • A later response indicates that the explanation provided makes more sense to the original poster.

Areas of Agreement / Disagreement

Participants appear to agree on the application of the chain rule to derive the expression for the derivative, but the initial confusion about the 2ax term indicates some uncertainty that was clarified through discussion.

Contextual Notes

The discussion does not delve into specific assumptions or limitations regarding the application of the chain rule or the definitions of the functions involved.

wahaj
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The derivative of sin(ax²) is 2axcos(ax²). Why is that? I understand the cos(ax²) part of the derivative but I don't understand where the 2ax is coming from.
 
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Do you know the chain rule?
 
wahaj said:
The derivative of sin(ax²) is 2axcos(ax²). Why is that? I understand the cos(ax²) part of the derivative but I don't understand where the 2ax is coming from.


You can google "Chain rule". It says that

$$\left(f(g(x)\right)'=f'(g(x))\cdot g'(x)$$

In the present case we have [itex]\,f(x)=\sin x\,\,\,,\,\,\,g(x)=ax^2\,[/itex]

DonAntonio
 
that makes a lot more sense. thanks
 

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