Question #8c1a5

1 Answer
Nov 26, 2017

dy/dx=-1/(2e) and -e/2

Explanation:

.

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