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What is the derivative of # x^2 x e^-x#?

Answer 1

Ezekiel Alexander

#3x^2e^-x-x^3e^-x#

Given: #x^2xe^-x#.

#=x^3e^-x#

We use the product rule, which states that,

#d/dx(xy)=yd/dx(x)+xd/dx(y)#

#:.dy/dx(x^3e^-x)=e^-xd/dx(x^3)+x^3d/dx(e^-x)#

#=e^-x*3x^2+x^3*-e^-x#

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

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Answer 2

Charles Arocha

To find the derivative of ( x^2 \cdot e^{-x} ), you can use the product rule.

The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), the derivative of their product is given by ( u'(x) \cdot v(x) + u(x) \cdot v'(x) ).

Let ( u(x) = x^2 ) and ( v(x) = e^{-x} ).

The derivative of ( u(x) = x^2 ) is ( u'(x) = 2x ).

The derivative of ( v(x) = e^{-x} ) is ( v'(x) = -e^{-x} ).

Now, applying the product rule:

( \frac{d}{dx} (x^2 \cdot e^{-x}) = u'(x) \cdot v(x) + u(x) \cdot v'(x) )

( = (2x) \cdot (e^{-x}) + (x^2) \cdot (-e^{-x}) )

( = 2x \cdot e^{-x} - x^2 \cdot e^{-x} )

So, the derivative of ( x^2 \cdot e^{-x} ) is ( 2x \cdot e^{-x} - x^2 \cdot e^{-x} ).

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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.