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How do you find the derivative of # x^2 * e^-x#?

Answer 1

Charles Anctil

#d/dx(x^2*e^-x)=2xe^-x-x^2e^-x#

This problem requires use of the product rule, which states: #d/dx(uv)=u'v+uv'# Where #u# and #v# are functions of #x#.

In our case, #u=x^2->u'=2x# and #v=e^(-x)->v'=-e^(-x)#. Thus #d/dx(x^2*e^-x)=(2x)(e^-x)+(x^2)(-e^-x)# #=2xe^-x-x^2e^-x#

We could simplify this a little further by, say, pulling out an #xe^-x#: #d/dx(x^2*e^-x)=xe^-x(2-x)#

Either form is correct.

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Answer 2

Paisley Johnson

To find the derivative of ( x^2 \cdot e^{-x} ), you can use the product rule of differentiation. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their product is given by:

[ (u \cdot v)' = u' \cdot v + u \cdot v' ]

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

[ u'(x) = 2x ] [ v'(x) = -e^{-x} ]

Now apply the product rule:

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

[ = 2x \cdot e^{-x} - x^2 \cdot e^{-x} ]

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

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