The slope of the tangent line is the value of the first derivative.

f(x) = 1/(x+1) = (x+1)^(-1)

f'(x) = -1(x+1)^(-2)

So at x=0,

m = f'(0) = -1(0+1)^(-2) = -1 /(1)^2 = -1

Using the point-slope form of the line, and the point (0,1):

(y -y1) = m(x -x1)

(y -1) = -1(x -0)

y -1 = -x

y = -x +1 ---------------the tangent line.

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If two lines are parallel, then they have the same slope.

f(x) = 1 /sqrt(x) = (x)^(-1/2)

f'(x) = (-1/2)x^(-3/2)

f'(x) = -1 /[2sqrt(x^3)] -------(i)

Slope of the line x + 2y-6=-

What?

Suppose the line is x + 2y-6 = 0

Transform it into the form y = mx +b,

x +2y -6 = 0

2y = -x +6

y = -(1/2)x +3

So, m = -1/2

Then, m = f'(x) = -1/2

We need the point of tangency so that we can use the point-slope form.

-1/2 = -1/[2sqrt(x^3)]

Divide both sides by (-1/2),

1 = 1/sqrt(x^3)

sqrt(x^3) = 1

Square both sides,

x^3 = 1

x = 1 ------at point of tangency.

What is the y there?

y = f(x) = 1/sqrt(x)

y = 1/sqrt(1)

y = 1

So, point of tangency is (1,1)

Then, with m = -1/2, and (1,1):

(y -1) = (-1/2)(x -1)

y = (-1/2)x +1/2 +1

y = -(1/2)x +3/2 ---------------the tangent line.