Why Does Differentiating y=sin(pi*x) Result in y'=cos(pi*x)*pi?

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Homework Help Overview

The discussion revolves around the differentiation of the function y=sin(pi*x) and the resulting expression for its derivative. The subject area pertains to calculus, specifically the application of the chain rule in differentiation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of the chain rule in the context of differentiating a trigonometric function with a composite argument. Questions arise regarding the correct identification of the outer and inner functions in this differentiation process.

Discussion Status

The discussion appears to be progressing with participants confirming their understanding of the chain rule and its application to the problem. Some guidance on the differentiation process has been provided, and there is a shared acknowledgment of the correct identification of functions involved.

Contextual Notes

Participants are working within the framework of calculus rules, specifically focusing on the chain rule for derivatives. There is an implicit understanding of the need for clarity in function composition during differentiation.

h_k331
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When y=sin(pi*x), why does y'=cos(pi*x)*pi, not y'=cos(pi*x)?

hk
 
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It's the chain rule. when the argument of a trig function is a FUNCTION of x, you have take the derivative of the agrgument. So, in general

(f(g(x)))'= f'(g(x))*g'(x)
 
So than in this case y=f(u)=sinu and u=g(x)=pi*x, correct?
 
h_k331 said:
So than in this case y=f(u)=sinu and u=g(x)=pi*x, correct?
That is correct.
 
Thanks guys, I appreciate it.

hk
 

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