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How do you differentiate #g(x) = (2e^(x^2) + 4e^x) ( 2x + 2x^2)# using the product rule?

Answer 1

Cameron Alexander

The product rule states that for a function #y=f(x)g(x)#, #(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#.

We'll also need chain rule, which indicates that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

#(dg(x))/(dx)=(4xe^x+4e^x)(2x+2x^2)+(2e^(x^2)+4e^x)(2+4x)#

#(dg(x))/(dx)=(4e^x(x+1))(2x(x+1))+(2e^x(e^x+2))(2+4x)#

#(dg(x))/(dx)=(x+1)(4e^x+2x)+(2e^x(e^x+2))(2+4x)#

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Answer 2

Emily Ammerman

To differentiate ( g(x) = (2e^{x^2} + 4e^x)(2x + 2x^2) ) using the product rule, follow these steps:

Let ( u = 2e^{x^2} + 4e^x ) and ( v = 2x + 2x^2 ).
Find ( u' ) and ( v' ), which are the derivatives of ( u ) and ( v ) with respect to ( x ), respectively.
Apply the product rule ( (uv)' = u'v + uv' ) to find the derivative of the product function ( g(x) ).

( u' = (4xe^{x^2}) + (4e^x) )

( v' = 2 + 4x )

Using the product rule:

( g'(x) = (2e^{x^2} + 4e^x)(2 + 4x) + (4xe^{x^2} + 4e^x)(2x + 2x^2) )

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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.