How do you find the derivative of # (x^(1/2))(e^(-x)) #?

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Answer 1 · 2016-10-02T11:18:13

#e^{-x} [ (1-2x)/(2sqrt(x)) ]#

The rule for deriving a product is

#(f*g)' = f'g+fg'#

In your case, we have

#f(x) = x^{1/2} = sqrt(x)#

#f'(x) = 1/2 x^{-1/2} = 1/(2sqrt(x))#

#g(x) = e^{-x}#

#g'(x) = e^{-x} * d/dx (-x) = e^{-x}(-1) = -e^{-x}#

I used the power rule #d/dx x^n = nx^{n-1}# for #f'(x)# and the composite rule #d/dx f(g(x)) = f'(g(x))*g'(x)# for g'(x)#

Now that we have all the elements, let's plug everything into the formula:

#f'g+fg' = 1/(2sqrt(x))*e^{-x}+ sqrt(x)(-e^{-x})#

Which we can rearrange factoring #e^{-x}#:

#e^{-x} [1/(2sqrt(x))-sqrt(x)]#

and again, if you prefer,

#e^{-x} [ (1-2x)/(2sqrt(x)) ]#