Let's start by differentiating #1/sqrt(x - 1)#.
We will use a combination of the chain rule and the quotient rule.
We need to first find the derivative of #sqrt(x - 1)#. Let #y = sqrt(u)# and #u = x - 1#.
#y' = 1/u^(1/2) xx 1 = 1/(2(x- 1)^(1/2)#
We can now apply the quotient rule.
#f'(x) = (0 xx sqrt(x- 1) - 1/(2(x - 1)^(1/2)) xx 1)/(sqrt(x - 1))^2#
#f'(x) = (-1/(2(x - 1)^(1/2)))/(x - 1)#
#f'(x) = -1/(2(x- 1)^(1/2)(x - 1)#
#f'(x) = -1/(2(x- 1)^(3/2)#
Next, we need to find the slope of the line #x + 2y + 7 = 0#. In the form #y = mx + b#, the slope is given by the parameter #m#.
#2y = -7 - x#
#y =-7/2 - 1/2x#
#:.# The slope is #-1/2#. Since the tangent is parallel, the tangent will have an equal slope to this line. The slope of the tangent is given by evaluating the point #x = a# into the derivative.
#-1/2 = -1/(2(x - 1)^(3/2)#
#-1(2)(x - 1)^(3/2) = -2#
#(x- 1)^(3/2) = 1#
#((x - 1)^(3/2))^(2/3) = 1^(2/3)#
#x- 1 = 1#
#x = 2#
Now , all we need to do is determine the y-coordinate of that point.
#y = 1/sqrt(2 - 1) = 1/sqrt(1) = 1/1 = 1#
By point-slope form, the equation of the tangent is as follows.
#y - y_1 = m(x - x_1)#
#y - 1 = -1/2(x- 2)#
#y - 1 = -1/2x + 1#
#y = -1/2x + 2#
Hopefully this helps!
