Prove that the derivative of an odd function is even

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SUMMARY

The derivative of an odd function is proven to be an even function using the limit definition of the derivative and the chain rule. An odd function satisfies the condition f(-x) = -f(x) for all x. By applying the chain rule to differentiate f(-x), it can be shown that the resulting derivative, f'(x), is even, meaning f'(-x) = f'(x). This establishes a clear relationship between the properties of odd functions and their derivatives.

PREREQUISITES
  • Understanding of odd and even functions
  • Familiarity with the limit definition of the derivative
  • Knowledge of the chain rule in calculus
  • Basic proficiency in calculus concepts
NEXT STEPS
  • Study the limit definition of the derivative in detail
  • Explore the properties of even and odd functions further
  • Practice applying the chain rule in various differentiation problems
  • Investigate the implications of derivatives in higher-order functions
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in understanding the relationship between function properties and their derivatives.

NWeid1
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A function f is an even function is f(-x)=f(x) for all x and is an odd function is f(-x)=-f(x) for all x. Prove that the derivative of an odd function is even and the derivative of an even function is off. I get what even and odd functions are but I'm not sure how to rigorously prove this. Thanks.
 
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NWeid1 said:
A function f is an even function is f(-x)=f(x) for all x and is an odd function is f(-x)=-f(x) for all x. Prove that the derivative of an odd function is even and the derivative of an even function is off. I get what even and odd functions are but I'm not sure how to rigorously prove this. Thanks.

One way to do this is to use the limit definition of the derivative.

PS - don't delete the parts of the posting template.
 
Or, and I think simpler, use the chain rule to differentiate f(-x) with respect to x.
 

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