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Calculus/The Slope of a Tangent Line at a Point

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Expert: - 4/28/2011


Question
Hello, I am taking a calculas class from a nearby city over the internet, but have been given next to no resources and am having trouble figuring out how to solve for the slope of a tangent line in a function other than x^2.
The question I am currently working on and have been having difficulty answering is the slope of the tangent line of F(x)=root(x+2) at the point (1,1)
So far I have worked out the secant line to (root(h-1))/h
Please help.

Answer
The slope of a line is the value of the derivative.
If F(x) = sqrt(x+2), that is the same as F(x) = (x+2)^0.5.
To find the slope, we need to find the derivative.
If F(x) = x^n, F'(x) = nx^(n-1).
Thus, the derivative is F'(x) = 0.5/(x+2)^0.5.
That is the slope of a function at a point.

To pass through (1,1), given the point on the curve is (x0,y0),
The slope should be (y0-1)/(x0-1).

Since the slope is the derivative, the equation to solve would be
(y0-1)/(x0-1) = 0.5/(x+2)^0.5 where y0 = (x0+2)^0.5.

I would put in what y0 was into the equation,
square both sides of the equation,
and then solve for x.

Given that x-value, the line can be found.

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