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What is the second derivative of #xe^(3x)#?

Answer 1

Samantha Adkison

The answer is: #y''=6e^(3x)+9xe^3x#

#y'=1*e^(3x)+x*e^(3x)*3=e^(3x)+3xe^(3x)#

#y''=e^(3x)*3+3(1*e^(3x)+x*e^(3x)*3)=#

#=3e^(3x)+3e^(3x)+9xe^(3x)=6e^(3x)+9xe^3x#.

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Answer 2

Axel Alvarez

To find the second derivative of ( xe^{3x} ), we first need to find the first derivative and then differentiate it again.

Given ( y = xe^{3x} ), the first derivative is:

[ \frac{dy}{dx} = e^{3x} + 3xe^{3x} ]

Now, we differentiate this result to find the second derivative:

[ \frac{d^2y}{dx^2} = \frac{d}{dx} (e^{3x} + 3xe^{3x}) ] [ \frac{d^2y}{dx^2} = 3e^{3x} + 3e^{3x} + 9xe^{3x} ] [ \frac{d^2y}{dx^2} = 6e^{3x} + 9xe^{3x} ]

So, the second derivative of ( xe^{3x} ) is ( 6e^{3x} + 9xe^{3x} ).

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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.