How do you find the derivative of #f(x)=xe^(-(x^2)/2)#

Answer 1

You can use the Product Rule where:

if #f(x)=g(x)*h(x)#

#f'(x)=g'(x)h(x)+g(x)h'(x)#

in your case #g(x)=x# but #h(x)=e^(-x^2/2)# which needs the Chain Rule to be derived (here you derive the #e# as it is and multiply times the derivative of the exponent).

So you get: #g(x)=x# #g'(x)=1# #h(x)=e^(-x^2/2)# #h'(x)=e^(-x^2/2)(-2x/2)#

and finally your derivative:

#f'(x)=1e^(-x^2/2)+xe^(-x^2/2)(-2x/2)=# #=e^(-x^2/2)(1-x^2)#

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Answer 2

Ezra Adams

To find the derivative of ( f(x) = x e^{-\frac{x^2}{2}} ), you can use the product rule and the chain rule. The derivative is:

[ f'(x) = e^{-\frac{x^2}{2}} + x \cdot \left(-\frac{2x}{2}\right) e^{-\frac{x^2}{2}} ] [ f'(x) = e^{-\frac{x^2}{2}} - x^2 e^{-\frac{x^2}{2}} ] [ f'(x) = (1 - x^2) e^{-\frac{x^2}{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.