Question

Find y'(x) Given y(x)=5-2(x+1)e^-1/2x (x is bigger or equal to 0)

Answer 1

\[y=5-2(x+1)e^{\frac{-1}{2}x}\] is this the function?

Answer 2

yes please

Answer 3

using the product rule,i can go as far as ;
-2(e^-1\2x)+5-2(x+1)(-1/2e^-1/2x)
the answer is (x-1)e^-1/2x.
my problem is i cant figure out how they got to the answer?
CAN ANYONE HELP!

Answer 4

yeah i will do it

Answer 5

\[y=5-2xe^{-\frac{1}{2}x}-2e^{-\frac{1}{2}x}\]\[y'=-2e^{-\frac{1}{2}x}+(-2xe^{-\frac{1}{2}x} \times \frac{-1}{2})+e^{-\frac{1}{2}x}\]\[=-2e^{-\frac{1}{2}x}+xe^{-\frac{1}{2}x}+e^{-\frac{1}{2}x}\]\[=-e^{-\frac{1}{2}x}+xe^{-\frac{1}{2}x}\]now factor out the e^(1/2)x
\[=e^{-\frac{1}{2}x}(-1+x)\]\[=(x-1)e^{-\frac{1}{2}x}\]