.
y=(1+x^(1/2))/e^((x^(1/2))
y=(1+x^(1/2))e^-(x^(1/2))
Since this is a product of two functions we will use the product rule:
If y=f(x)*g(x)
dy/dx=f'(x)*g(x)+g'(x)*f(x)
In our problem:
f(x)=1+sqrtx=1+x^(1/2)
g(x)=1/e^(sqrtx)=e^-(x^(1/2))
f'(x)=1/2x^(-1/2)=1/2*1/x^(1/2)=1/(2sqrtx)
To take the derivative of g(x), let u=-x^(1/2)
du=-1/(2sqrtx) (we took the derivative of the positive of this function above)
Then g(x)=e^u
g'(x)=e^udu=e^-(x^(1/2))*-1/(2sqrtx)=(1/e^(x^(1/2)))(-1/(2sqrtx))
g'(x)=-1/(2e^(sqrtx)sqrtx
Therefore:
dy/dx=1/(2sqrtx)(1/e^(sqrtx))+(-1/(2e^sqrtxsqrtx))(1+sqrtx)
dy/dx=(1/(2e^sqrtxsqrtx))-((1+sqrtx)/(2e^sqrtxsqrtx))=(1-1-sqrtx)/(2e^sqrtxsqrtx)=-sqrtx/(2e^sqrtxsqrtx)=-cancelcolor(red)sqrtx/(2e^sqrtxcancelcolor(red)sqrtx)=
dy/dx=-1/(2e^sqrtx)
At x=1
dy/dx=-1/(2e^(+-1)
dy/dx=-1/(2e)
and
dy/dx=-1/(2e^-1)=-1/(2/e)=-e/2