How do you differentiate #g(x) = 2xe^(2x)# using the product rule?

Answer 1

#f=2x,g=e^(2x) ->f'=2,g'=2e^(2x)#
#g'(x)=fg'+gf'=4xe^(2x)+2e^(2x)#

Before adding any of the derivatives to the product rule, separate the products into f and g.

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Answer 2

To differentiate ( g(x) = 2xe^{2x} ) using the product rule, we follow these steps:

  1. Identify the two functions being multiplied together: ( u(x) = 2x ) and ( v(x) = e^{2x} ).
  2. Apply the product rule formula: ( (uv)' = u'v + uv' ).
  3. Calculate the derivatives of ( u(x) ) and ( v(x) ): ( u'(x) = 2 ) and ( v'(x) = 2e^{2x} ).
  4. Substitute these values into the product rule formula: ( (2x)'e^{2x} + 2x(e^{2x})' ).
  5. Simplify: ( 2e^{2x} + 2x(2e^{2x}) ).
  6. Further simplify: ( 2e^{2x} + 4xe^{2x} ).

So, the derivative of ( g(x) = 2xe^{2x} ) using the product rule is ( g'(x) = 2e^{2x} + 4xe^{2x} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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