Question

1. find the slope of the tangent to the curve y=1/sqroot(x) at x=a.
2. find the equation of the line at (4,1/2)

Answer 1

The slope of the tangent to a curve f(x) at x = a is just f’(a),
ie. the derivative evaluated at x = a.

Once you know the slope of the tangent line, and you
want to find the equation of the tangent line itself,
you’d probably want to use the point-slope form
y – y1 = m(x – x1) since you know that your tangent line passes through a point (x1, y1).

That automatically gives you the equation of the line.

Here y = 1/x^(1/2) = x^(-1/2). So y’ = (-1/2) x^(-3/2).
Then y’(a) = (-1/2) a^(-3/2) will be the slope of the tangent line at x = a

Then the equation of the tangent line is y – 1/2 = (-1/2) * (4)^(-3/2) (x – 4).

Notice that we plugged in the derivative at x = 4 for the slope m.

Simplify to get the final answer.