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How do you differentiate #g(y) =(1-x^2)e^(3x) # using the product rule?

Answer 1

Luke Alvarez

#(dg)/(dx)=e^(3x)(3-2x-3x^2)#

Product rule states that #d/(dx)f(x)g(x)=(df)/(dx)xxg(x)+f(x)xx(dg)/(dx)#

Thus, in the specified function,

#(dg)/(dx)=(d(1-x^2))/(dx)xxe^(3x)+(1-x^2)xx(d(e^(3x)))/(dx)# or

= #(-2x)xxe^(3x)+(1-x^2)3e^(3x)# or

= #-2xe^(3x)+3e^(3x)(1-x^2)# or

= #e^(3x)(-2x+3(1-x^2))# or

= #e^(3x)(3-2x-3x^2)#

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Answer 2

Christopher Agresta

To differentiate ( g(y) = (1 - x^2)e^{3x} ) using the product rule:

Identify the functions being multiplied: ( f(x) = 1 - x^2 ) and ( h(x) = e^{3x} ).
Apply the product rule: ( g'(x) = f'(x)h(x) + f(x)h'(x) ).
Find the derivatives of ( f(x) ) and ( h(x) ).
- ( f'(x) = -2x )
- ( h'(x) = 3e^{3x} ).
Substitute the derivatives and functions into the product rule formula: ( g'(x) = (-2x)(e^{3x}) + (1 - x^2)(3e^{3x}) ).
Simplify the expression: ( g'(x) = -2xe^{3x} + 3e^{3x} - 3x^2e^{3x} ).
Combine like terms if necessary, yielding the final result.

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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.