# How do you find the instantaneous rate of change of #y=4x^3+2x-3# at #x=2#?

So:

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Are you applying the definition or do you have rules for differentiation?

Important: If you are using a definition, it's probably one of the two below. You'll get the same answer either way, but I'm not sure where you are in your calculus study.

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First Solution:

Option 2:

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To find the instantaneous rate of change of ( y = 4x^3 + 2x - 3 ) at ( x = 2 ), you can first find the derivative of the function with respect to ( x ), which represents the rate of change of ( y ) with respect to ( x ) at any given point. Then, evaluate the derivative at ( x = 2 ) to find the instantaneous rate of change.

The derivative of the function ( y = 4x^3 + 2x - 3 ) is ( y' = 12x^2 + 2 ).

Evaluating the derivative at ( x = 2 ), we get: ( y'(2) = 12(2)^2 + 2 ) ( y'(2) = 12(4) + 2 ) ( y'(2) = 48 + 2 ) ( y'(2) = 50 )

So, the instantaneous rate of change of ( y = 4x^3 + 2x - 3 ) at ( x = 2 ) is ( 50 ).

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