Arturo O. answered • 06/02/16

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First, let us find the slope of x+2y+6=0:

y = -(1/2)x - 3

Slope is m = -1/2

Next, set dy/dx of the given function equal to m. That will be the slope of a line parallel to x+2y+6=0:

y = 1 / sqrt(x)

dy/dx = -(1/2) / x^(3/2) = m = -1/2

x^(3/2) = 1, so x = 1, and f(1) = 1 / sqrt(1) = 1 (note x>0, so we can neglect the negative square root)

So far, the tangent point is (1,1) and the slope of the tangent line is m = -1/2. Now put this in form y=mx+b and find b:

1 = -(1/2)(1) + b, so b = 3/2

Final equation of tangent line at (1,1):

y = -(1/2)x + 3/2

I think you had some difficulty evaluating the derivative of 1/sqrt(x). Just use the form:

y = x^k with k = -1/2

dy/dx = kx^(k-1) = -(1/2)x^(-1/2 - 1) = -(1/2) / x^(3/2)