If P(x) = g^2(x), then P'(3) = ?

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SUMMARY

The discussion centers on finding the derivative P'(3) for the function P(x) = g^2(x). Participants confirm that the correct approach involves applying the chain rule, leading to the derivative expression P'(x) = 2g(x)g'(x). Substituting x = 3 results in P'(3) = 2g(3)g'(3). The conversation highlights the importance of understanding both the chain rule and product rule in calculus for correctly deriving functions.

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  • Understanding of calculus concepts, specifically derivatives.
  • Familiarity with the chain rule and product rule in differentiation.
  • Knowledge of function notation and evaluation.
  • Basic understanding of the notation g^2(x) as g(x) multiplied by itself.
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  • Study the application of the chain rule in calculus.
  • Review the product rule for differentiation in detail.
  • Practice problems involving derivatives of composite functions.
  • Explore examples of functions defined as squares or products of other functions.
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Students learning calculus, educators teaching differentiation techniques, and anyone seeking to improve their understanding of derivative calculations involving composite functions.

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Homework Statement


If P(x) = g^2(x), then P'(3) =


Homework Equations





The Attempt at a Solution


I am not quite sure what g^2(x) means...
But my assumption is do the derivative of g^2(x), so it becomes 2g(x), then put the 3 in for x?
so the final answer will be 2g(3) ?
It looks weird to me, so I am not sure if I am doing it correctly or not.

Please advise, thank you.
 
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Almost! you need to apply chain rule!
 
Also, your inclination that g2(x) = g(x)*g(x) is correct.
 
To sutupidmath,
Thank you!
So it should be 2g(x)g'(x), thus the final answer for that question would be P'(3) = 2g(3)g'(3).

To Mark44
I got confused a little there because I thought the final answer was different from 2g(x) (which was the wrong answer anyways). Now I redid the problem using product rule instead for g(x)g(x), and my derivative turned out to be g'(x)g(x) + g(x)g'(x), which is the same as 2g(x)g'(x) anyways :D

Thank you so much to both of you.
 

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