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Find y'(x) Given y(x)=5-2(x+1)e^-1/2x (x is bigger or equal to 0)
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\[y=5-2(x+1)e^{\frac{-1}{2}x}\] is this the function?
yes please
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!
yeah i will do it
\[y=5-2xe^{-\frac{1}{2}x}-2e^{-\frac{1}{2}x}\]\[y'=-2e^{-\frac{1}{2}x}+(-2xe^{-\frac{1}{2}x} \times \frac{-1}{2})+e^{-\frac{1}{2}x}\]\[=-2e^{-\frac{1}{2}x}+xe^{-\frac{1}{2}x}+e^{-\frac{1}{2}x}\]\[=-e^{-\frac{1}{2}x}+xe^{-\frac{1}{2}x}\]now factor out the e^(1/2)x \[=e^{-\frac{1}{2}x}(-1+x)\]\[=(x-1)e^{-\frac{1}{2}x}\]
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