First we need to differentiate the function in order to find the gradient at x=2.
If we rewrite x/(sqrt(2x+2)) as x*(2x+2)^(-1/2) we can use the product rule and the chain rule.
The product rule:
dy/dx (a*b)= b*(da)/dx + a*(db)/dx
d/dx(x)=1
d/dx(2x+2)^(-1/2)=-1/2(2x+2)^(-3/2)*(2)=-(2x+2)^(-3/2)
dy/dxx*(2x+2)^(-1/2)=(2x+2)^(-1/2) +x * (-(2x+2)^(-3/2))
->(1)/((2x+2)^(1/2))-(x)/(2x+2)^(3/2)=((2x+2)-x)/((2x+2)^(3/2))=(x+2)/((2x+2)^(3/2))=(x+2)/((2x+2)(2x+2)^(1/2))=1/2(x+2)/((x+1)sqrt(2x+2)
f^'(x)=1/2(x+2)/((x+1)sqrt(2x+2)
Gradient at x=2
1/2(2+2)/((2+1)sqrt(2(2)+2))=4/(6sqrt(6))=sqrt(6)/9
For the normal line:
sqrt(6)/9=-(3sqrt(6))/2
Point on the line:
Plugging 2 into original function:
2/(sqrt(2(2)+2))=2/(sqrt(6))=(sqrt(6))/3
(2color(white)(8) , (sqrt(6))/3)
y-(sqrt(6))/3=-(3sqrt(6))/2(x-2)
->y=-(3sqrt(6))/2x+(6sqrt(6))/2+(sqrt(6))/3=-(3sqrt(6))/2x+(10sqrt(6))/3
Normal line:
y==-(3sqrt(6))/2x+(10sqrt(6))/3
Or:
6y=-9sqrt(6)*x+20sqrt(6)
Plot:
![enter image source here]()
