.
#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#
